The Inverse Galois Problem: the Rigidity Method

Home , Galois extension, Inverse Galois problem

The Inverse Galois Problem: The Rigidity Method

Amin Saied Department of Mathematics, Imperial College London, SW7 2AZ, UK

CID:00508639

June 24, 2011

Supervisor: Professor Martin Liebeck Abstract

The Inverse Galois Problem asks which finite groups occur as Galois groups of extensions of Q, and is still an open problem. This project explores rigidity, a powerful method used to show that a given group G occurs as a Galois group over Q. It presents the relevant theory required to understand this approach, including Riemann’s Existence Theorem and Hilbert’s Irreducibility Theorem, and applies them to obtain specific conditions on the group G in question, which, if satisfied say that G occurs as a Galois group over the rationals.

These ideas are then applied to a number of groups. Amongst other results it is explicitly shown that Sn,An and PSL2(q) for q 6≡ ±1 mod 24 occur as Galois groups over Q. The project culminates in a ﬁnal example, highlighting the power of the rigidity method, as the Monster group is realised as a Galois group over Q. Contents

1 Introduction 2 1.1 Preliminary Results ...... 2 1.2 A Motivating Example ...... 4

2 Riemann’s Existence Theorem 5 2.1 The Fundamental group ...... 5 2.2 Galois Coverings ...... 7 2.3 Coverings of the Punctured Sphere ...... 8

3 Hilbertian Fields 13 3.1 Regular Extensions ...... 13 3.2 Hilbertian Fields ...... 14 3.3 Galois Groups Over Hilbertian Fields ...... 19 3.3.1 Sn as a Galois group over k ...... 20

4 Rigidity 21 4.1 Laurent Series Fields ...... 21 4.2 Finite Extensions of Λ ...... 22 4.3 Branch Points ...... 24 4.4 Rigid Ramiﬁcation Types ...... 26 4.5 Structure Constants ...... 28

5 The Rigidity Criteria 33 5.1 Descent of the Base Field ...... 33 5.2 Rigidity Criteria ...... 38

6 Applications of Rigidity 41 6.1 Sn and An as Galois groups over Q ...... 41 6.2 PSL2(q) for q 6≡ ±1 mod 24 as a Galois group over Q ...... 43 6.3 M12 as Galois group over Q ...... 46 2 2 ab 6.4 The Ree Groups G2(q ) as Galois groups over Q ...... 47 6.5 Realising the Monster over Q ...... 49

7 Conclusion 51

References 52

A GAP Code 53

B Characters for M12 55

1 Chapter 1

Introduction

By associating a certain group to a given polynomial, Evariste´ Galois was able to prove that there is no general solution by radicals to a polynomial equation of degree 5 or higher. His approach provided not only a beautiful answer to this classical problem, but also oﬀered a great insight into why it is possible to solve polynomial equations of degree less than 5 in general.

Galois’ idea was to associate to a polynomial p(x) a certain group Gal(p(x)) which permutes the roots of p(x). In general, given a polynomial with rational coefficients, the Galois group is defined on a certain field extension E/Q, where E is the splitting field of p(x). It is here that a natural question arises.

Inverse Galois Problem: Which groups can be realised as Galois groups over Q? The problem was ﬁrst approached by Hilbert, in 1892, and remains unsolved today. This project develops the most successful approach to the question thus far, namely the concept of rigidity. Introduced by John Thompson in 1984 in his breakthrough paper [11], this method has already been used to realise numerous families of groups including, and perhaps most notably, all but 2 of the 26 sporadic groups.

A striking feature of the material is the vast diversity of topics it draws from, including group theory, ﬁeld theory, Galois theory, algebraic topology, Riemann surface theory and number theory. This project develops the relevant theorems from these areas and pulls them together to form a strict group theoretic condition, rigidity, which if satisﬁed in a group G, along with some other conditions, will guarantee a positive solution to the Inverse Galois Problem for G.

Some background material is set up and some basic results from Galois theory are stated. Proofs, where not provided, and a more thorough description of the background can be found in [2] and [10].

1.1 Preliminary Results

A field extension E/K is algebraic if every element of E is algebraic over F . An algebraic field extension is said to be normal if every irreducible polynomial f(x) ∈ K[x] which has a root in L, has all its roots in L. An algebraic field extension E/K is separable if every x ∈ L has a separable minimal polynomial over K, that is, the minimal polynomial has distinct roots. A Galois extension is an algebraic field extension E/K which is normal and separable. The degree of an extension E/K is the dimension of E as a vector space over K and is denoted

2 |E : K|. An extension is said to be ﬁnite if it has ﬁnite degree. Moreover for nested extensions E/K, K/F , the degree of E/F obeys the tower law |E : F | = |E : K||K : F | as in [10].

Definition 1.1.1. Let E/K be a finite Galois extension. Then the Galois group is defined as

Gal(E/K) := {σ ∈ Aut(E): σ(x) = x ∀ x ∈ K}

This next basic result is fundamental to Galois theory and will be used throughout, often without reference.

Lemma 1.1.2. Let f(x) ∈ K[x] and let E/K be a Galois extension. If α ∈ E is a root of f(x) and σ ∈ Gal(E/K), then σ(α) ∈ E is a root of f(x).

n n Proof. Let f(x) = anx + ··· + a0 with ai ∈ K. Now f(α) = 0 so anα + ··· + a0 = 0. Now applying σ ∈ Gal(E/K) gives

n n σ(anα + ··· + a0) = anσ(α) + ··· + an = 0 because σ ﬁxes K pointwise. Hence f(σ(α)) = 0

Theorem 1.1.3. (The Primitive Element Theorem) Let E/K be a ﬁnite Galois extension. Then ∃ α ∈ E such that E = K[α].

I now quote the main theorem of Galois theory, from [2].

Theorem 1.1.4. Let E/K be a finite Galois extension with Galois group G = Gal(E/K). For an intermediate field F such that K ⊆ F ⊆ E let λ(F ) be the subgroup of G leaving F fixed. Then λ is a one-to-one map of the set of all such intermediate fields F onto the set of subgroups of G with the following properties:

1. λ(F ) = Gal(E/F )

2. F = EGalE/F = Eλ(F ) where EH denotes the ﬁxed ﬁeld of H ≤ G in E.

3. For H ≤ G, λ(EH ) = H

4. |E : F | = |λ(F )|, |F : K| = |G : λF )|

5. F is a normal extension of K if and only if λ(F ) is a normal subgroup of G. When this is the case . Gal(F/K) = Gal(E/K) Gal(E/F )

This correspondence between subgroups of Gal(E/K) and intermediate ﬁelds F between E and K is referred to as the Galois correspondence, and it too will be used throughout this project. I now quote another classic result from [10].

Theorem 1.1.5. (Artin’s Theorem) If G is a finite group of automorphisms of a field E and if the fixed field of G in E is K then E/K is a finite Galois extension with Gal(E/K) = G

3 1.2 A Motivating Example

How does one find a group as a Galois group? In the true nature of the original idea of Galois theory, one might think to look at polynomials f(x) ∈ Q[x]. To find the Galois group of a polynomial one must find its splitting field Ef , which is the smallest field extension of Q containing the roots of f [10]. Lets consider the dihedral group of the square D8. Can it be realised as a Galois group over Q?

Consider the irreducible polynomial

4 2 f(x) = x − 6x + 2 ∈ Q[x]

In general the splitting field Ef of an irreducible polynomial of degree n can have degree up to n! over Q, and thus the Galois group is a subgroup of Sn [10]. Therefore in our case the splitting field Ef has degree dividing 24 (by Lagrange). This is why it was a good idea to look at a polynomial of degree 4, as D8 is the unique subgroup of S4 of order 8. Proceed by finding√ 2 2 the roots of f(x) by first setting y = x and finding the√ roots of y − 6y + 2, which are 3 ± 7. An extension of degree 2 has been created, namely Q( 7). Now find the roots of f(x) to be q √ q √ ±α := 3 + 7 ± β := 3 − 7 √ √ Hence the splitting field of f is E = (α, β). Notice that αβ = 9 − 7 = 2 ∈/ (α). Now f Q √ Q √ 2 2 p β ∈ Q(α) ⇐⇒ αβ ∈ Q(α). Therefore β∈ / Q(α). Each equation x = 3+ 7 and x = 3 − 7 produces an extension of degree 2, and so we have that Ef is a degree 8 extension of Q. Therefore |Gal(f)| = |Gal(Ef /Q)| = 8, and since D8 is the unique subgroup of S4 of order 8 we have that 4 2 ∼ Gal(x − 6x + 2) = D8

So by a clever choice of polynomial one was able to realise D8 as a Galois group over Q. This approach essentially tries to construct a ﬁeld extension which yields the desired group as its Galois group. However, this will become increasing more diﬃcult as the group in question becomes more complicated. For instance, how would one use this technique to realise the Monster group. The mind boggles.

This motivates the need to develop a more powerful technique. This will be rigidity. Rigidity takes the opposite approach, that is, rather than creating a field extension which yields the desired group G, rigidity is a condition on the group itself, which if satisfied, along with some other conditions, ensures the existence of a field extension over Q with Galois group G.A fundamental concept in this vein is Riemann’s Existence Theorem.

4 Chapter 2

Riemann’s Existence Theorem

Riemann’s Existence Theorem is one of the foundations on which rigidity is built. Eventually it will ensure the existence of a field extension of C(x), the complex function field, whose Galois group is G, where G satisfies some specific conditions. It is based on coverings of the punctured Riemann sphere, and this chapter begins by setting up the relevant background in algebraic topology (for a more thorough development see [3]). It goes on to explore Galois coverings and ends with the proof of a topological version of the theorem.

2.1 The Fundamental group

Recall some basic notions from algebraic topology. Let S be a topological space. A path from p to q is a continuous map γ : I → S with γ(0) = p, γ(1) = q. C(S, p, q) denotes the set of all paths in S from p to q. Two paths γ0, γ1 ∈ C(S, p, q) are homotopic if ∃ a continuous map Γ: I × I → S with Γ(0, t) = γ0(t), Γ(1, t) = γ1(t) for all t ∈ I and Γ(s, 0) = p, Γ(s, 1) = q for all s ∈ I. Homotopy of paths is an equivalence relation, and we denote the equivalence class of γ by [γ]. The set of homotopy classes of closed paths (i.e. with same start point and end point) based at p in S is denoted π1(S, p).

Consider two paths g, h : I → S such that g(1) = h(0). Define the product of these two paths by 1 g(2t) t ∈ [0, 2 ] hg : I → S t 7→ 1 h(2t − 1) t ∈ [ 2 , 1] The path γ has a natural inverse under this product γinv : I → S defined by γinv(t) = γ(1 − t), which traverses γ backwards. Moreover, this product induces a product of homotopy classes. We are now able to define the fundamental group.

Deﬁnition 2.1.1. The set π1(S, p) of homotopy classes based at p in S together with the above product is a group, called the fundamental group of S based at p.

The groups π1(S, p) and π1(S, q) are isomorphic if ∃ a path γ ∈ C(S, p, q). Then an isomorphism is given by [a] 7→ [γaγinv]..

A (topological) manifold is topological space that is second countable, Hausdorﬀ and locally Euclidean. From now on we consider S, a manifold. A surjective map f : R → S is called a covering if ∀ p ∈ S ∃ an open neighbourhood U such that f −1(U) is a disjoint union of open sets in R, each of which is mapped homeomorphically onto U by f. Such a U is called

5 an admissible neighbourhood. It follows from this deﬁnition that R, the covering space, is locally Euclidean. Indeed each admissible neighbourhood U is locally Eulclidean, and f −1(U) is homeomorphic to U. Let f : R → S be a covering, and γ a path in S. Then γe is a lift of γ if f ◦ γe = γ. Recall the homotopy lifting property from [3], Theorem 2.1.2. Let f : R → S be a covering. For each a ∈ f −1(p) ∃! lift of γ to R with starting point a. Moreover, if two paths in S are homotopic, then their lifts are homotopic if they have the same start point.

−1 Define the fiber of p as f (p). The fundamental group π1(S, p) acts naturally on the fiber −1 −1 f (p) thusly: [γ] ∈ π1(S, p) sends a ∈ f (p) to b, the endpoint of the unique lift of γ starting at a. Suppose S is connected. Then this action is transitive if and only if R is connected. Now recall from [3] two results on coverings

Lemma 2.1.3. Let f : R → S be a covering and suppose S is connected. Let R1 be a connected −1 component of R. Then the restriction of f to R1 is a covering. Moreover, if f (p) ⊂ R1 then R1 = R is connected. Lemma 2.1.4. Let f : R → S and h : T → S be two coverings with R and T connected. Let p = f(a)=h(b) with a ∈ R and b ∈ T . Suppose [γ]a = a ⇐⇒ [γ]b = b. Then ∃ a homeomorphism α : R → T with h ◦ α = f and α(a) = b.

Corollary 2.1.5. If π1(S, p) is trivial then each covering f : R → S with R connected is a homeomorphism.

Proof. Let h = Id : S → S. Then f : R → S and h : S → S are two coverings and R and S are connected by hypothesis. Now since π1(S, p) is trivial we can use 2.1.4 to get a homeomorphism α : R → S. But h ◦ α = f i.e. α = f so indeed f is a homeomorphism.

For a covering f : R → S the isomorphisms α : R → R are called deck transformations, and they form a group under composition, denoted Deck(f) or Deck(R/S) if f is understood [3]. The group Deck(f) acts naturally on the ﬁber f −1(p) thusly: for a ∈ f −1(p), α ∈ Deck(f) then α(a) ∈ f −1(p).

For a covering f : R → S, a path γ from p to q and a ∈ f −1(p) denote by γa the endpoint of the lift of γ with initial point a. This gives us a bijection between the fibers f −1(p) and f −1(q), and moreover, this bijection commutes with the action of Deck(f), that is α(γa) =γ α(a) [12]. For γ a loop at p we have [γ] ∈ π1(S, p). As a result we have that Deck(f) commutes with the action of π1(S, p) Lemma 2.1.6. Let R be connected. If α ∈ Deck(f) fixes a point a ∈ R then α = Id. Moreover, if G is a subgroup of Deck(f) that acts transitively on a fiber f −1(p) then G = Deck(f).

Proof. Let γe be a path from a to b, some b ∈ R. Then γ = f ◦ γe, and γe is the lift of γ with initial point a and end point γa = b. Now α ∈ Deck(f) ﬁxes a by assumption, that is α(a) = a. But then α(b) = α(γa) =γ α(a) =γ a = b. Now b was arbitrary, so α = Id.

Consider again a ∈ f −1(p). For each α ∈ Deck(f) there is β ∈ G with β(α(b)) = b But then by above βα = Id ⇒ α = β−1 ∈ G.

6 2.2 Galois Coverings

If f : R → S is a covering then the cardinality of f −1(p) is locally constant over S. Hence if S is connected then this cardinality is constant on all S. The cardinality is then called the number of sheets or the degree of f [3].

Deﬁnition 2.2.1. A covering f : R → S is called a Galois covering if R is connected and Deck(f) acts transitively on each ﬁber f −1(p), p ∈ S.

Lemma 2.2.2. Let f : R → S be a Galois covering and let H := Deck(f). The number of sheets of f is equal to the order of H.

−1 Proof. Exhibit a bijection from H to a general ﬁber. Let a ∈ f (p). Consider map θa : H → f −1(p) by α 7→ α(a). −1 −1 θa injective: α, β ∈ H. Suppose α(a) = β(a) then β α(a) = a. By 2.1.6 then β α = Id ⇒ β = α therefore θa is injective. θa surjective: f is a Galois covering, so acts transitively on the each ﬁber. Therefore θa is clearly surjective.

Intuitively, if the degree of f is n then each ﬁber has n components, and H := Deck(f) permutes them transitively among themselves.

Proposition 2.2.3. Let f : R → S be a Galois covering. For a ∈ R, p ∈ S there is a unique surjective homomorphism

Φa : π1(S, p) → Deck(f)

[γ] 7→ Φa([γ]) where Φa([γ]) : [γ]a 7→ a. Here [γ]a is the endpoint of the lift of γ starting at a.

−1 −1 Proof. π1(S, P ) acts transitively on f (p) so the map Ψa : π1(S, p) → f (p) by [γ] 7→ [γ]a is −1 surjective. But composition of surjections is surjective, so θa ◦ Ψa : π1(S, p) → H is surjective −1 −1 (as θa is a bijection). Now let Φa :[γ] 7→ ((θa ◦ Ψa)[γ]) . Since π1(S, p) commutes with the action of H, it is trivial but tedious to check that Φa is indeed a homomorphism with

7 0 the required property. Suppose that Φa is another surjective homomorphism with the same 0 properties. That is Φa([γ]) : [γ]a 7→ a. But then −1 0 Φa([γ]) ◦ Φa([γ]) : a 7→ a −1 0 0 thus by 2.1.6, Φa([γ]) ◦ Φa([γ]) = Id ⇒ Φa([γ]) = Φa([γ]). This shows that indeed Φa is unique.

Example: The Punctured Disc

Let K(r) := {z ∈ C : 0 < |z| < r} be a disc of radius r minus the origin. It suﬃces to consider K = K(1) as all K(r) are homeomorphic√ to K. Let H := {z ∈ C : Re(z) < 0} the open left half z plane. Let i denote the ﬁxed choice of −1 in C. Consider the map f∞ : H → K by z 7→ e . z z0 0 Now e = e ⇐⇒ z = λm(z) := z + 2πmi for m ∈ Z.

Let z ∈ H and let V = B(z), 0 < ≤ 1, the open disc of radius about z contained in H. Then V ∩ λm(V ) = ∅ for m ∈ Z. Deﬁne U := f∞(V ), open in K, and for each m ∈ Z the map f∞ restricted to λm(V ) is a homeomorphism onto U, thus U is an admissable neighbourhood for q = f∞(z) and thus f∞ : H → K is a covering.

Now by construction each λm is a deck transformation of f∞. Moreover it is clear that the −1 group G = {λm : m ∈ Z} acts transitively on the ﬁber f∞ (p) hence f∞ is a Galois covering and Deck(f) = G (by 2.1.6).

Now consider the associated map Φa : π1(K, p) → Deck(f∞). I claim it is an isomorphism. Indeed, suppose [g] ∈ ker(Φa). Lift g to a loop at a. But π1(H, a) = {1} is trivial [3]. Hence the lift of g is homotopic to a constant path and thus g is homotopic to a constant path at p i.e. [g] = 1. So Φa is injective. It is by construction surjective and thus indeed is an isomorphism.

2πit Deﬁne a path eh : I → H by eh(t) = a + 2πit. Let h = f∞ ◦ eh, then h(t) = pe . Now [h]a = eh(1) = a + 2πi. Thus Φa([h]) : a + 2πi 7→ a ⇒ Φa([h]) : z 7→ z − 2πi. Therefore π1(K, p) is inﬁnite cyclic and is generated by [h].

2.3 Coverings of the Punctured Sphere

1 1 2 Let P := C ∪ {∞}. P is a topological space. We can view it as R ∪ { a point} and so see that 2 1 it is homeomorphic to S [3]. Let P be a finite subset of P . It is our goal to show that finite 1 Galois coverings of P \P correspond to finite Galois extensions of C(x).

Define the open disc D(p, r) of radius r about p by {z ∈ C : |z − p| < r} p ∈ C D(p, r) = 1 {z ∈ C : |z| > r } ∪ {∞} p = ∞ 1 Let f : R → P \P be a finite Galois covering. Let D = D(p, r) for some p ∈ P such that D ∗ ∗ 1 doesn’t contain any other elements of P . Define D = D \ p. Then D ⊂ P \P . We can now define

∗ κp : D → K(r) z − p p 6= ∞ z 7→ 1 z p = ∞

8 −1 ∗ Let E be a connected component of f (D ), and deﬁne fE = κp ◦ f|E. This is a covering −1 ∗ ∗ f|E : E → K(r). Indeed the covering f restricts to a covering f (D ) → D . So composition −1 ∗ with the homeomorphism κp gives a covering f (D ) → K(r) which restricts to a covering E → K(r) by 2.1.3. E is called a circular component of level r over p. Let 0 < rb < r. There is a bijection between cirular components E of level r and circular components Eb of level rb over p. Indeed, there is precisely one E ⊂ E, and f is the restriction of f to E. This allows us to b Eb E b deﬁne an equivalence relation on the set of circular components over p: E ∼ E0 if E ⊂ E0 or E0 ⊂ E. The equivalence classes of ∼ are called the ideal points of R over p.

−1 ∗ Lemma 2.3.1. Deck(f) permutes the components E of f (D ) transitively. If HE is the stabiliser of E in H := Deck(f) then restriction to E gives an isomorphism HE → Deck(fE). Proof. H acts on f −1(D∗) and hence permutes the components E. Let h ∈ H. If h maps one point of E to E0, some other component, then h maps all of E to E0, since components are pairwise disjoint. So if h maps some point in E to some other point in E then h ∈ HE i.e. ∗ −1 stabilises E. Let q ∈ D and consider the set FE = f (q) ∩ E, a ﬁber of the f|E. Since −1 f is a Galois covering, H acts transitively on the ﬁber f (q) any two points of FE can be mapped into each other by an h ∈ H ,and so this h must be in HE. So we have that HE acts transitively on FE. We have a homomorphism HE → Deck(fE),, which is injective by 2.1.6 and whose image is a subgroup of Deck(fE) acting transitively on FE. Thus by 2.1.6, again, each subgroup is all of Deck(fE) and thus restriction gives the required isomorphism.

As a result, and by the example of the punctured disc, HE is cyclic. Let hE be the distinguished generator of HE.

∗ ∗ ∗ −1 2πit ∗ ∗ Let p ∈ D , p = κp(p ) and λ(t) = κp (pe ), a loop in D based at p . Let b ∈ R and 1 ∗ deﬁne q0 = f(b). Further, let δ be a path in P \P joining q0 to p . Lift δ to δe via f with initial point b. Then δe has endpoint b∗ in some component E. Let λe be the lift of λ via f with initial ∗ 2πit ∗ point hE(b ). Then λe is the lift of the path pe via fE with initial point hE(b ). Thus the endpoint of λe is b∗ [12].

0 −1 Let h ∈ H := Deck(f) and E = h(E) another connected component. Then hhEh = hE0 where hE is the distinguished generator of HE. Therefore the stabilisers hE form a conjugacy class Cp of Deck(f) which depends only on p and f. Indeed h ◦ λe is the lift of λ via f with ∗ 0 ∗ initial point h(hE(b )) ∈ h(E) = E . Now h ◦ λe has endpoint h(b ) and hE0 maps the endpoint ∗ ∗ −1 −1 ∗ ∗ of h ◦ λe to the initial point. So hE0 (h(b )) = h(hE(b )) ⇒ hE h hE0 h(b ) = b therefore by −1 2.1.6 is identity i.e. hh h = h 0 . If E ⊂ E then h = h so indeed the class doesn’t depend E E b Eb E on the radius of the disc [12].

inv 1 1 Deﬁne γ = δ λδ be a closed path in P \P based at q0, and the map Φb : π1(P \P, q0) → H sends [γ] to an element in Cp. Indeed the path (hE ◦ δ) is the lift of δ starting at hE(b). This ends at b therefore Φb([γ]) = hE ∈ Cp.

−1 ∗ Deﬁnition 2.3.2. Let Cp be the conjugacy class of stabilisers of the components E of f (D ). Then, given f, Cp depends only on p. Let e be the common order of the elements of Cp.

Then e is equal to the degree of the covering fE : E → K(r) for any component E, and in particular, Cp = {1} if and only if fE is a homeomorphism [12].

9 Ideal Points 1 Let f : R → P \P . The ideal points are the equivalence classes of ∼ deﬁned on the circular components by E ∼ E0 if E ⊂ E0 or E0 ⊂ E. So the number of ideal is equal to the number of componants E of f −1(D∗). Now D∗ = D \ p, so the circular components E are homeomorphic to a disc minus the center, the ideal points can be thought of as the missing centers. Since Deck(f) permutes the components transitively 2.3.1 the number of ideal points over p is ≤ |Deck(f)|. Infact it is intuitively clear that the the number of ideal points is less than or equal to the number of sheets of the covering.

1 Let f : R → P \P be a finite Galois covering. Let R be the disjoint union of R and all ideal points over all p ∈ P . Define a topology on R by V ⊂ R is open if V ∩ R is open in R and the following holds: for each ideal point π ∈ V there is an E ∈ π with E ⊂ V . This makes R into a connected compact Hausdorff space [12]. Note that if π is an ideal point then for each E ∈ π the set E ∪ {π} is an open neighbourhood of π. Moreover, each neighbourhood of π contains such a set.

1 1 Now we can extend f : R → P \P to a continuous surjective map f : R → P by setting f(π) = p where π is an ideal point over p. Further, each α ∈ Deck(f) extends uniquely to a homeomorphism α : R → R by f ◦ α = f. [ Lemma 2.3.3. Let X be a topological space with open subsets Xj, j ∈ J, such that X = Xj. j Then each path δ : I → X is homotopic to a product of ﬁnitely many paths δν with each δν in some Xj.

Proof. Consider the preimages of Xj under δ. They clearly cover I. I is a closed bounded subset −1 −1 of R so by Hiene-Borel is compact, thus there exists a ﬁnite subcover. Choose δ (X1), . . . , δ (Xs) with s minimal. Order the Iν such that inf(Iν) < inf(Iν+1). Then Iν ∩ Iν+1 6= ∅ by minimality of s. Let tν be in this overlap. Then we have 0 = t0 < t1 < ··· < ts−1 < ts = 1 and δ([tν, tν+1]) lies entirely within Xj for some j. Thus after a suitable re-parametrisation the result is obtained.

Let p1, . . . , pn ∈ C distinct and set S = C\{p1, . . . , pn}. It is possible to choose a point such that the straight line between this point and the pi hit no other pj. Let q0 ∈ S be such a iν point. Can write pi = q0 + ρie i . Relabel the pi so that the vi are increasing. Here of course νi represents the angle the line from q0 to pi makes with the real line. Now divide the complex plane up, into connected components Si, with rays, M1,...,Mn emanating from q0 such that each pi lies in Si. Let Di be a disc about pi entirely contained within Si. Deﬁne paths γi starting from q0, travelling along the straight line from q0 to pi until it hits the boundary ∂Di, traversing the boundary once clockwise, and travelling back to q0 along this same line.

10 Theorem 2.3.4. [γ1],..., [γn] generate π1(S, q0).

0 0 Proof. Enlarge Si to Si = {z ∈ C : dist(a, Si) < } such that Si ∩ Dj 6= ∅ for i 6= j. Then 0 0 the Si form an open cover of C and thus the Si\pi form an open cover of S. Now consider a general loop in S based at q0. By 2.3.3 γ can be decomposed into a product of paths δν with ν = 1, . . . , s such that each δ is a path in T := S0 \{p }. Let κ be the straight line path from ν ν iν iν ν inv q0 to δν lying in Tν ∩ Tν−1. Now the path ων = κν+1δνκν is a loop at p0 (note, let κs+1 be the constant path at q0). Now γ is homotopic to the product of the ων. Also, Tν is homeomorphic to the punctured disc K which has inﬁnite cyclic fundamental group as calculated earlier, and m is generated by the γi with i = iν. So each ων is homotopic to some γi , m ∈ Z. But then γ is a product of these ω , so is homotopic to a product of γm, so indeed the γ generate the ν iν i fundamental group.

I now quote a result from [12]

Theorem 2.3.5. Let S = C\{p1, . . . , pn}. Let G be a group with generators g1, . . . , gn. Then −1 ∃ a Galois covering f : R → S, an isomorphism θ : Deck(f) → G and a point b ∈ f (q0) such that the composition of θ with the surjection Φb : πq(S, q0) → Deck(f) from 2.2.3 maps [γi] to ∼ gi for i = 1, . . . , n. Now G = Deck(f). 1 If G is ﬁnite then f : R → P \{p1, . . . , pn, ∞} is a ﬁnite Galois covering, and the associated −1 conjugacy classes Ci = Cpi and C∞ of G as in 2.3.1 are such that gi ∈ Ci and (g1 ··· gn) ∈ C∞ for i = 1, . . . , n.

The proof of this theorem is omitted due its length. It can be found in [12]. As a sketch, consider the line from q0 to pi. Let Li be the continuation of this line emanating from pi. Let Q = C\{L1,...,Ln}. Then, roughly speaking, R is obtained by glueing sheets Q × {g}, which are indexed by g ∈ G, along the rays Li × g. We see that G acts naturally on the indices, and this gives the permutation of the sheets required for the identiﬁcation with Deck(f).

Lemma 2.3.6. The fundamental group π1(S, q0) where S = C\{p1, . . . , pn} is freely generated by [γ1],..., [γn].

Proof. Let G be the free group generated by g1, . . . , gn. Then by 2.3.5 there exists a homomorphism π1(S, q0) → G sending [γi] to gi. But there is also a homomorphism from G → π1(S, q0) sending gi to [γi]. Composition both ways is identity as it ﬁxes either all gi which generate G, or

11 all [γi] which generate π1(S, q0). Hence G is isomorphic to π1(S, q0) and under this isomorphism gi ↔ [γi] i.e. π1(S, q0) is generated by [γ1],..., [γn].

2 2 It follows that since S minus a point is homeomorphic to C, that S \{p1, . . . , pn+1} is 0 0 homeomorphic to S = C\{p1, . . . , pn} and thus is free of rank n.

The spaces S2 and S2\{p} both have trivial fundamental group, the latter is seen by its asso- 2 ciation with R or C as mentioned above [3].

Our next goal is to use Theorem 2.3.4 and 2.3.5 to prove a topological version of Riemann’s 1 existence theorem. First, consider a finite Galois covering f : R → P \P , where P is a finite 1 subset of P . Let H = Deck(f) and as in definition 2.3.2, let Cp be the class of H associated 0 to p. Consider the points giving rise to non-trivial conjugacy classes P := {p ∈ P : Cp 6= {1}}. 0 0 0 0 Thus we have a triple (H,P , (Cp)p∈P 0 ). We say two such triples (H,P, C), (H ,P , C ) are 0 0 0 equivalent if P = P and there is an isomorphism from H to H mapping Cp to Cp.This defines an equivalence relation on such triples.

Deﬁnition 2.3.7. The (topological) ramiﬁcation type of the covering f is the equivalence 0 0 class of the triple (H,P , (Cp)p∈P 0 ) denoted [H,P , (Cp)p∈P 0 ].

1 The ramiﬁcation type arises from a well deﬁned triple depending only on f : R → P \P . 0 0 Indeed, H = Deck(f) and P ⊂ P such that Cp 6= {1} ⇐⇒ p ∈ P . Given such a Galois covering another is obtained from a change of coordinates in the following way. Consider a homeomorphism,

1 1 g : P −→ P

1 sending z ∈ C to z − p0 if p0 ∈ C, and ∞ 7→ ∞. Or sending z ∈ C\{0} to z , ∞ 7→ 0 and 0 7→ ∞

1 Then fe := g ◦ f : R → P \g(P ) is another finite Galois covering with Deck(fe) = Deck(f). 0 Hence the ramification type of fe is [H, g(P ), (Cg−1p)p∈g(P 0)]. Theorem 2.3.8. (Topological) Riemann’s Existence Theorem Let T = [G, P, (Kp)p∈P ] be a ramification type. Let r = |P | and label the elements of P as 1 p1, . . . , pr. Then there exists a finite Galois covering of P \P of ramification type T if and only if there exist generators g1, . . . , gr of G with g1 ··· gr = 1 and gi ∈ Kpi . Proof. We can ensure, possibly by a change of coordinates, that ∞ ∈ P , and can label the elements such that pr = ∞. 1 Firstly, suppose f : R → P \P is a finite Galois covering of ramification type T . Choose 1 inv q0 ∈ P \P and paths γ1, . . . , γr−1 as for Theorem 2.3.4. Let γr = (γ1 · γr−1) . Consider the −1 1 fiber f (q0), and fix some point b in this fiber. Then Φb : π1(P \P, q0) → G is a surjective homomorphism, as in 2.2.3. Let gi := Φb([γi]). Then by Theorem 2.3.4 the gi generate G and by Theorem 2.3.5, gi ∈ Cpi = Ci. Finally, γ1 ··· γr = 1 by the definition of γr. Thus g1 ··· gr = 1 as Φb is a homomorphism.

Conversely, suppose ∃ generators g1, . . . , gr of G with g1 ··· gr = 1 and gi ∈ Kpi . Apply Theorem 1 2.3.5 which gives the existence of a ﬁnite Galois covering f : R → P \P and an isomorphism between Deck(f) and G such that gi lies in the class of Cpi associated to pi for i = 1, . . . , r − 1. −1 ∼ gr = (g1 ··· gr−1) ∈ C∞ = Cpr . Hence Deck(f) = G, and Cp = Kp ∀ p ∈ P , so indeed f has ramiﬁcation type T as needed.

12 Chapter 3

Hilbertian Fields

It was Hilbert who first systematically approached the Inverse Galois Problem, and his ideas still form a crucial role in the functioning of the rigidity method. His main idea allows a Galois group over Q(x), the rational function field, to be realised as a Galois group over Q. This chapter explores the theory of so called hilbertian fields and concludes with the statement of Hilbert’s Irreducibility Thorem.

3.1 Regular Extensions

0 Given a FG-extension k /k and x = (x1, . . . , xm) with the xi’s algebraically independent, what can we say about k0(x)/k(x)?

Lemma 3.1.1. Given FG-extension k0/k and x as above, then k0(x)/k(x) is a FG-extension and Gal(k0(x)/k(x)) =∼ Gal(k0/k).

0 0 Proof. Extend action of G on k to k (x) by g ∈ G then g(xi) = xi for i = 1, . . . m. G clearly ﬁxes k(x) and hence k0(x)/k(x) is a FG-extension with Galois group G.

Remark 3.1.2. By the Galois correspondence we see that any intermediate ﬁeld between k0(x) and k(x) is of the form k00(x) and [k00(x): k(x)] = [k00 : k]

Deﬁnition 3.1.3. L is said to be regular over k if k is algebraically closed in L. Say L/k is regular or, sometimes, geometric.

Lemma 3.1.4. Let k¯ be an algebraic closure of k. If f(x, y) ∈ k(x)[y] is an irreducible polyno- . mial over k(x), and if K = k(x)[y] (f) is the corresponding ﬁeld extension of k(x), then

K is regular over k ⇐⇒ f is irreducible over k¯(x)

Proof. (⇐) Let bk be the algebraic closure of k in K. In other words, K is regular over k iﬀ bk = k.

Let f an irreducible poly over k(x) and let α be a root of f.

We have a homomorphism

ϕ : k(x)[y] −→ k(x)(α) h(y) 7→ h(α)

13 . . where k(x)(α) =∼ k(x)[y] ker(ϕ) = k(x)[y] (f) = K

Then α satisﬁes a polynomial fb(y) ∈ bk(x)[y] of degree [K : bk(x)] and fb divides f. Indeed [K : k(x)] [K : bk(x)] = ≤ [K : k(x)] (by tower law) [bk(x): k(x)] Hence deg(fb) ≤ deg(f)

Therefore if bk 6= k then f not irreducible over bk(x)[y] and hence not irreducible over k¯(x)[y].

(⇒) Assume K is regular over k, (then k = bk). Let k0 be a FG-extension of k and deﬁne K0 to be the composite of K and k0(x) in an algebraic closure of k(x).

Then by 3.1.1 k0(x)/k(x) is a FG-extension. And by Remark 3.1.2. K ∩ k0(x) is of the form k00(x) for k00 an intermediate ﬁeld between k0 and k.

Now k00 ⊂ bk = k hence k00 = k i.e. K ∩ k0(x) = k(x). Since k0(x)/k(x) is Galois, it follows that [K0 : k0(x)] = [K : k(x)]

3.2 Hilbertian Fields

Definition 3.2.1. A field k is called hilbertian if for each irreducible polynomial f(x, y) ∈ k[x, y] with deg(f) ≥ 1 there are infinitely many b ∈ k such that f(b, y) ∈ k[y] is irreducible.

The polynomial fb(y) = f(b, y) is called a specialisation. Remark 3.2.2. Let k be a ﬁeld. Let b be algebraic over k. That is, b satisﬁes

n n−1 anb + an−1b + ··· + a0 = 0

n n−1 n−2 n−1 Define a monic polynomial g(y) = y + (a0an + a1an y + ··· an−1y ) such that g(ban) = 0. Lemma 3.2.3. Let k be a field, and let f(x, y) ∈ k[x, y] be separable when viewed over as a polynomial in one variable f(y) ∈ k(x)[y]. Then there are only finitely many b ∈ k such that f(b, y) that are not separable. Proof. By remark we may assume that f is monic in y. Viewed as a polynomial in one variable y f ∈ k(x)[y] is separable, therefore the discriminant Disc(f) is a non-zero element of k(x). Say Disc(f) = h(x). Then for b ∈ k fb(y) has discriminant h(b) ∈ k. This is zero if and only if b is a root of h, which only happens finitely often. Therefore the specialisation fb is not separable only finitely often.

Let D be an integral domain, and let F be its field of fractions. D is called a factorial domain if there exists P ⊂ A such that 0 ∈/ P and any b ∈ D can be written uniquely as a Y product b = u pn(p) with u a unit of D, and this product is finite (i.e. n(p) = 0 for all but p∈P finitely many p). The best known examples of factorial domains are Z and k[x] the polynomial ring, both of which are PIDs. It is known that all PIDs are factorial domains [8]. In fact, a polynomial ring in n variables is known to be factorial [5].

Let f(y) ∈ D[y] be a polynomial over the factorial domain D such that deg(f) ≥ 1. f(y) is irreducible in D[y] if and only if it is irreducible in F [y], and f(y) is called primitive if the gcd of all nonzero coeﬃcients is 1. For all g(y) ∈ F [y] ∃! d ∈ F such that dg(y) is primitive [5].

14 Lemma 3.2.4. Let f(x1, . . . , xs) ∈ k[x1, . . . , xs] be in a polynomial in s > 1 variables. Then f is irreducible if and only if when viewed as a polynomial in one variable fs(xs) ∈ k[x1, . . . , xs−1][xs], fs is irreducible and primitive.

Proof. Clearly if f(x1, . . . , xs) is irreducible then fs(xs) is irreducible and primitive. Con- versely, suppose fs(xs) is irreducible and primitive. If f = gh can view this as fs = gh for g, h ∈ k[x1, . . . , xs] then, WLOG, g ∈ k[x1, . . . , xs−1] else contradicts irreducibility of fs over k[x1, . . . , xs−1][xs]. f primitive ⇒ g a unit in k[x1, . . . , xs−1], which is a factorial domain. Hence f is irreducible when viewed as a polynomial in s variables.

Consider the finite Galois extension L/k of degree n with G = Gal(L/k). Let κ ⊆ k such that k is the field of fractions of κ i.e. k = (κ∗)−1κ. Let a generate L over k such that there exists f(y) ∈ κ[y] monic, satisfying a with deg(f) = n. Lemma 3.2.5. Suppose there exists A ⊂ L such that a ∈ A, A is invariant under G and the field generated by κ and A, κ[A] = κ[a]. Then there exists w ∈ κ such that for each ring homomorphism φ : κ → k0, k0 a field, φ(w) 6= 0 and φ extends to a map Φ: κ[a] → L0 where L0/k0 is a finite Galois extension with Galois group G0 = Gal(L0/k0). L0 is generated over k0 by a0 := Φ(a). Let f 0 be the polynomial obtained by applying φ to all the coefficients of f. Then f 0(a0) = 0. Furthermore, if f 0 is irreducible then there exists a unique isomorphism G → G0 sending g 7→ g0 such that Φ(g(x)) = g0(Φ(x)) (3.1) for all g ∈ G, x ∈ κ(a). Proof. L/k is a Galois extension ⇒ f(y) ∈ κ[y] is separable i.e. has no repeated roots. Hence the discriminant Disc(f) 6= 0 ∈ κ. Now φ(Disc(f)) = Disc(f 0). Consider φ : κ → k0 such that Disc(f 0) 6= 0 i.e. such that f 0(y) is separable. Extend φ : κ → k0 to a map ψ : κ[y] → k0[y] where ψ(y) = y. This map induces a homomorphism

. 0 . Ψ: κ[y] fκ[y] → k [y] f 0k0[y] since f 0 is obtained by applying φ to all the coefficients of f, so ψ(f(y)) = f 0(y) by definition. Here fκ[y] denotes the ideal generated by f in κ[y]. . There is a natural isomorphism ζ : κ[y] fκ[y] → κ[a] by h 7→ h(a). Now define

−1 0 . η := Ψ ◦ ζ : κ[a] → κ [y] (f 0) Here we are taking the quotient by (f 0), the ideal generated by f 0 in κ0[y]. A larger field is obtained by taking the quotient of κ0[y] by (g0), for g0 an irreducible factor of f 0. This field is 0 . naturally a field extension of κ [y] (f 0), and so composing with η there is an extension of φ to the map 0 . 0 . 0 κ[a] → κ [y] (f 0) → κ [y] (g0) = L and this is Φ. Need to show that L0 is a Galois extension of k0. Now G permutes the conjugates a1, . . . , an of a amongst themselves, and since by hypotheses A is invariant under G, all conjugates of a lie in A ⊂ κ[A] = κ[a]. Write

f(y) = (y − a1) ··· (y − an)

15 then applying φ to this get

0 0 0 f (y) = (y − a1) ··· (y − an)

0 0 0 0 0 0 where ai = φ(ai). Now f (a ) = 0, so a1, . . . , an are the conjugates of a , and so the conjugates of a0 all lie in L0 over k0, and so L0/k0 is a normal extension. But f 0 is separable, so therefore L0/k0 is separable i.e. L0/k0 is indeed a ﬁnite Galois extension.

0 0 0 Finally, suppose f is irreducible. It is separable, i.e. has distinct roots i.e. a1, . . . , an are 0 0 0 0 distinct. There is a unique gi ∈ G mapping a to ai by properties of Galois groups. Similarly 0 0 there is a unique gi ∈ G mapping a to ai. Hence there is a bijection G → G sending gi → gi. Need to show this is an isomorphism satisfying (3.1). Pick x ∈ κ[a]. That is, there is polynomial h(y) ∈ κ[y] such that x = h(a). Let h0(y) = Φ(h(y)) ∈ κ0[y]. Then

0 0 0 0 0 0 0 gi(Φ(x)) = gi(Φ(h(a))) = gi(h (a )) = h (ai) = Φ(h(ai)) = Φ(gi(h(a))) = Φ(gi(x))

0 0 0 0 This shows (3.1). Now if g1, g2 ∈ G then (g1g2) (a ) = (g1g2) (Φ(a)) = Φ((g1g2)a) = g1(Φ(g2(a))) = 0 0 0 0 0 g1g2(Φ(a )). Therefore the map G → G sending gi 7→ gi is a homomorphism, and it is a bijection, so it is an isomorphism, as required.

Remark 3.2.6. The condition that κ[A] = κ[a] in Lemma 3.2.5 can be removed. Indeed, x ∈ A Pn−1 i can be written x = i=1 bia with bi ∈ k. Since k is the field of fractions of κ, can choose c ∈ κ such that cbi ∈ κ. Recall that Disc(f) is a non-zero element of κ, as f(y) ∈ κ[y] is separable. −1 Define d = cDisc(f) and let κe = κ[d ]. Then cbi ∈ κ ⇒ cDisc(f)bi ∈ κ ⇒ dbi ∈ κ ⇒ bi ∈ κe. Therefore x ∈ κe[a] for all x ∈ A. So κe[A] = κe[a]. This is the case in Lemma 3.2.5. This next result forms the basis of the key theorem in this chapter, which in turn is a key concept in the theory of rigidity when applied to the Inverse Galois Problem. Lemma 3.2.7. Let L/k(x) be a finite Galois extension. Then there is a monic polynomial f(x, y) ∈ k[x, y] and a generator a of L over k(x) such that f(x, a) = 0. Moreover, for all but finitely many b ∈ k, if the specialised polynomial fb(y) = f(b, y) is irreducible in k[y] then k[y] . (fb) is Galois over k with Galois group isomorphic to Gal(L/k(x)).

Proof. a generates L over k(x), so a satisﬁes a polynomial f(y) ∈ k(x)[y]. Let a1, . . . , an be the conjugates of a over k(x). Can view f as a polynomial in two variables f ∈ k[x, y], and by Remark 3.2.2 we may assume f is monic in y. Write

f(y) = (y − a1) ··· (y − an) (3.2)

Now a is amongst the conjugates, hence f(a) = f(x, a) = 0.

For the last part let b ∈ k. There is a natural homomorphism

φb : k[x] → k h(x) 7→ h(b)

0 Now by Lemma 3.2.5 to φb : k[x] → k. Obtain f (y) by applying φb to the coefficients of f(y). 0 Then by definition of φb, f (y) = f(b, y) = fb(y). Lemma 3.2.5 says that if there exists w ∈ k[x] . k[y] such that φb(w) 6= 0 then the field (fb) is Galois over k with Galois group Gal(L/k(x)). Now φb(w) = w(b) is zero only when b is a root of w, i.e. only finitely often. This completes the proof.

16 Lemma 3.2.8. Let L/k(x) be a finite Galois extension, and let l be a finite extension of k such that l ⊂ L. Then if p(x, y) ∈ l[x, y] is irreducible as a polynomial in one variable y over l(x) such that the roots of p are contained in L, then for all but finitely many b ∈ k, if fb(y) ∈ k[y] is irreducible then pb(y) = p(b, y) ∈ l[y] is irreducible, where f is as in Lemma 3.2.7 Qr Proof. Write p(x, y) = p0(x) i=1(y − bi) with bi ∈ L, p0(x) ∈ l[x]. Let bi ∈ A ⊂ L containing a generator of L over k, say a (this is the A from Lemma 3.2.5). Recall Φ : κ[A] → L0. Let 0 bi = Φ(bi). Let A contain a generator of l over k. Then Φ is defined on l too. Now assuming 0 fb is irreducible, then Φ maps l isomorphically to a subfield of L . Then applying Φ to p(x, y) Qr 0 gives p(b, y) = p0(b) i=1(y − bi). p(x, y) is irreducible, hence separable as we are assuming k has characteristic 0. By Lemma 3.2.5 Gal(L/l(x)) is mapped onto Gal(L0/l). By properties of 0 Galois groups, Gal(L/l(x)) acts transitively on the bi, and so Gal(L /l) acts transitively on the 0 bi. By Lemma 3.2.3 there are only finitely many b ∈ k for which p(b, y) is not separable, and there are finitely many b such that p0(b) = 0, namely the roots of p0. So for all b ∈ k excluding 0 these finitely many, the polynomial pb(y) is separable, and Gal(L /l) acts transitively on the roots. Therefore pb is irreducible. Lemma 3.2.9. Let L/k(x) be a finite Galois extension. There are finitely many polynomials pi(x, y) ∈ k[x, y] which are irreducible when viewed as polynomials in one variable y over k(x) and such that for all but finitely many b ∈ k, if none of the pi(b, y) has a root in k(x) then fb(y) is irreducible in k[y], where f is as in Lemma 3.2.7.

Proof. By Lemma 3.2.7, f(x, y) ∈ k[x, y] is irreducible when viewed as a polynomial in one variable y over k(x), and f is of the form f(y) = (y −a1) ··· (y −an) where ai are the conjugates of a over k(x), where a is a generator of L over k(x), as in (3.2). Hence L/k(x) has degree n. Q Let I ⊂ {1, . . . , n}. Now i∈I (y − ai) ∈/ k(x)[y], as f(y) is irreducible over k(x). Hence it has 0 some coefficient cI ∈/ k(x). Recall from Lemma 3.2.5 the map Φ : κ[A] → L where the set A contained a and its conjugates. Let pI be the an irreducible polynomial satisfying cI over k(x). Now suppose that fb is not irreducible. Then there exists an I of the form described above such Q that i∈I (y − ai) lies in k[y]. Now d = Φ(cI ) is a coefficient of this polynomial, therefore d ∈ k. But now applying Φ to pI (x, ci) = 0 gives pi(b, d) = 0. Proposition 3.2.10. The field k is hilbertian if and only if, given a finite extension l/k, and irreducible polynomails pi(x, y) ∈ l(x)[y] for i = 1, . . . , m, there are infinitely many b ∈ k such that pi(b, y) are irreducible in l[y] for i = 1, . . . , m. Proof. (⇒) Let p1(x, y), . . . , pm(x, y) ∈ l(x)[y]. Form a set A of all the roots of these m polynomials in an algebraically closed field containing l(x). Let L/l(x) be a finite extension such that A ⊂ L and L is Galois over k(x). Now apply Lemma 3.2.8. (⇐) Clear.

Proposition 3.2.11. The ﬁeld k is hilbertian if and only if, for any irreducible pi(x, y) ∈ k(x)[y] for i = 1, . . . , m, there are inﬁnitely many b ∈ k such that none of the specialised polynomials pi(b, y) ∈ k[y] has a root in k Proof. (⇒) Follows from the above proposition and Lemma 3.2.9. (⇐) is again clear.

Propositions 3.2.10 and 3.2.11 are essentially equivalent definitions for hilbertian fields, and I will use them interchangeably. In this set up, hilbertian fields k have a particularly useful property, that if a finite group G occurs as a Galois group over k(x1, . . . , xn), then G also occurs as a Galois group over k. I first need to establish some basic properties of hilbertian fields. The

17 properties established so far apply to polynomials in two variables, but this next remark on the Kronecker specialisation, provides a bridge to polynomials in s > 1 variables.

Remark 3.2.12. Let f(x1, . . . , xs) be a polynomial in s > 1 variables over a hilberitan field k. The Kronecker specialisation of f is defined by first letting d = df be an integer greater then any power of any variable xi occurring in f, and then by specialising thusly

d d2 ds−2 Sdf(x, y) = f(x, y, y , y , . . . , y ) (3.3) Q a polynomial in two variables x and y. Now, write Sdf(x, y) = g(x) i gi(x, y) with gi irreducible over k[x, y] and g ∈ k[x]. k is hilbertian i.e. there are inﬁnitely many b ∈ k such that gi(b, y) remains irreducible over k[y].

Let Sdf(x, y) = g(x)H1(x, y)H2(x, y). Then there exists unique polynomials eh1(x1, . . . , xs), eh2(x1, . . . , xs) such that Sdeh1 = gH1 and Sdeh2 = H2 and the highest power of x2, . . . , xs occurring is less then d [12].

Lemma 3.2.13. Let k be a hilbertian ﬁeld with f(x1, . . . , xs) ∈ k[x1, . . . , xs] irreducible over k. Then there are inﬁnitely many b ∈ k such that f(b, x2, . . . , xs) remains irreducible over k. Q Proof. Let Sdf(x, y) = g(x) i gi(x, y) be the Kronecker specialisation of f. Suppose for a contradiction that f(b, x2, . . . , xs) = h1(x2, . . . , xs)h2(x2, . . . , xs) is reducible, and consider their Kronecker specialisations Sdh1(y) and Sdh2(y). Let b be such that the gi(b, y) remain irreducible over k[y] from Remark 3.2.12, and also such that b is not a root of g(x). Then Q Sdf(b, y) = Sdh1(y)Sdh2(y) = g(b) i gi(b, y). So Sdh1(y), Sdh2(y) are products of the irreducible factors gi(b, y). Let H1(x, y),H2(x, y) be the product of the corresponding gi(x, y). Hence write Sdf(x, y) = g(x)H1(x, y)H2(x, y).

There exists unique polynomials eh1(x1, . . . , xs), eh2(x1, . . . , xs) such that Sdeh1 = gH1 and Sdeh2 = H2 as in Remark 3.2.12. Let fe = eh1eh2. If the highest power of fe occuring is less than d, then fe = f. But f is irreducible by hypothesis, hence this is a contradiction. Therefore the highest power of fe occurring is greater than or equal to d. Hence fe contains a monomial

i2 is f0(x1)x2 ··· xs (3.4) such that ij ≥ d some j ∈ [2, . . . , s]. Now Sdf(b, y) = Sdh1(y)Sdh2(y) and Sdf(x, y) = g(x)H1(x, y)H2(x, y) = Sdeh1Sdeh2. Therefore, ehi(b, x2, . . . , xs) is a scalar multiple of hi(x2, . . . , xs), for i = 1, 2 ⇒ fe(b, x2, . . . , xs) is a scalar multiple of f(b, x1, . . . , xs). Hence the monomial must vanish upon evaluation at b ∈ k else contradicting that d is greater then any power of any variable xi in f. This means that f0(b) = 0 from (3.4). But this clearly happens only ﬁnitely often. Hence it is the case that there are inﬁnitely many b ∈ k such that f(b, x2, ··· , xs) remains irreducible over k.

The following corollary shows that it is possible to specialise numerous variables in an irreducible polynomial f(x1, . . . , xs) ∈ k[x1, . . . , xs] in a hilbertian ﬁeld k and still have an irreducible polynomial.

Corollary 3.2.14. Let k be a hilbertian ﬁeld and f(x1, . . . , xs) an irreducible polynomial in s > 1 variables. Then for any p(x1, . . . , xs−1) ∈ k[x1, . . . , xs−1] there are b1, . . . , bs−1 ∈ k such that p(b1, . . . , bs−1) 6= 0 and f(b1, . . . , bs−1, xs) is irreducible.

18 Proof. By induction on s. Let s = 2, and suppose f(x1, x2) irreducible. Then by Lemma 3.2.13 there exists infinitely many b ∈ k such that f(b, x2) is irreducible over k. Hence statement holds. Fix s > 2 and suppose statement true for s − 1. View p(x1, . . . , xs−1) ∈ k[x1, . . . , xs−1] as a polynomial p1(x2, . . . , xs−1) ∈ k(x1)[x2, . . . , xs−1], with some coefficients polynomials in x1, say cj(x1). Then by Lemma 3.2.13 there exists b1 ∈ k such that fb1 (x2, . . . , x2) := f(b1, x2, . . . , xs) is irreducible and cj(b1) 6= 0 for some j. Then pb1 (x2, . . . , xs−1) := p(b1, x2, . . . , xs−1) is non-zero as some of its coefficients cj(b1) are non-zero. But now by the inductive hypothesis on pb1 a polynomial in s−2 variables, we have that there exists b2, . . . , bs−1 ∈ k such that pb1 (b2, . . . , bs−1) 6= 0 and fb1 (b2, . . . , bs) is irreducible. Therefore b1, . . . , bs−1 ∈ k works, and the inductive step is proved.

Lemma 3.2.15. Any ﬁnitely generated extension of a hilbertian ﬁeld is hilbertian.

Proof. Let l be a finite extension of k. Then for f ∈ l[x, y] an irreducible polynomial, then there are infinitely many b ∈ k and hence in l such that f(b, y) is irreducible in l[y], by definition (see Proposition 3.2.10. Hence l is hilbertian. Now consider a purely transcendental extension K = k(x1, . . . , xm). Let R = k[x1, . . . , xm], and f(x, y) ∈ K[x, y] irreducible in y. Then f is irreducible in y over K(x). Then WLOG f ∈ R[x, y], so by Lemma 3.2.4 f is irreducible as a polynomial in x1, . . . , xm, x, y. Hence by Lemma 3.2.13 there are infinitely many b ∈ k such that f(x1, . . . , xm, b, y). Therefore, when viewed as a polynomial in one variable y over R, and hence over K, f(x1, . . . , xm, b, y) = fe(y) ∈ R[y] ⊂ K[y] is irreducible. Therefore K is hilbertian.

Let K = k(a1, . . . , an) be a finitely generated extension of k. If x1, . . . , xm is a maximal algebraically independent subset of a1, . . . , an over k, then K is finite over the purely transcendental extension k(x1, . . . , xm). Indeed K is algebraic over k(x1, . . . , xm) and hence finite over k(x1, . . . , xm). Hence any finitely generated extension K of k is a finite extension of k or a finite extension of a purely transcendental extension of k. Either way K is hilbertian.

3.3 Galois Groups Over Hilbertian Fields

This section provides results that will prove invaluable later on, and forms one of the core arguments when using rigidity to realise groups as Galois groups over Q. Theorem 3.3.1. Let k be a hilbertian field and G a finite group. If G occurs as a Galois group over k(x1, . . . , xm) then G occurs as a Galois group over k. Proof. By induction on m. For m = 1 use Lemma 3.2.7. Let m > 1. Now k is hilbertian by hypothesis, therefore k(x1, . . . , xm−1) is a finitely generated extension of k and so by Lemma 3.2.15 is hilbertian. Then k(x1, . . . , xm) = k(x1, . . . , xm−1)(xm) so we reduce to the case m = 1 and use Lemma 3.2.7 again.

Definition 3.3.2. Let G be a finite group. Then G occurs regularly over k if there is a finite Galois extension L of k(x1, . . . , xm), regular over k with G = Gal(L/k(x1, . . . , xm)). Theorem 3.3.3. If G occurs regularly over k then G occurs regularly over every finitely generated extension l of k. Thus G is a Galois group over l if l is hilbertian.

Proof. G occurs regularly over k, say G = Gal(K/k(x1, . . . , xm)), and K is regular over k by deﬁnition. Then there is irreducible f ∈ k(x1, . . . , xm)[y] such that K is of the form . K = k(x1, . . . , xm)[y] (f)

19 with f(x1, . . . , xm, y) ∈ k(x1, . . . , xm)[y] irreducible over l(x1, . . . , xm) by Lemma 3.1.4. Con- . struct a ﬁeld extension of l(x1, . . . , xm) thusly, L = l(x1, . . . , xm)[y] (f). Now the roots of f over l(x1, . . . , xm) are contained in K already. Hence L/l(x1, . . . , xm) is a Galois extension, and we have Gal(L/l(x1, . . . , xm)) = Gal(K/k(x1, . . . , xm)). Finally, if l is hilbertian then G occurs as a Galois group over l by Theorem 3.3.1.

3.3.1 Sn as a Galois group over k

Consider the polynomial f(y) ∈ k(x1, . . . , xn)[y] deﬁned by

n n−1 f(y) = y + x1y + ··· + xn

Let a1, . . . , an be the roots of f(y). Then Sn acts naturally on the roots in the following way, let σ ∈ Sn then σ :(a1, . . . , an) 7→ (aσ(1), . . . , aσ(n))

This action extends to an automorphism σe : k(a1, . . . , an) 7→ k(a1, . . . , an). In this way we have constructed an action of Sn on k(a1, . . . , an). Consider the ﬁxed ﬁeld F of this action in k(a1, . . . , an). Then F contains k(x1, . . . , xn), and further |k(a1, . . . , an): F| = |Sn| = n!. But since a1, . . . , an are the roots of a polynomial of degree n over k(x1, . . . , xn) it follows that |k(a1, . . . , an)/k(x1, . . . , xn)| ≤ n!. Therefore F = k(x1, . . . , xn) and so by Artin’s theorem Gal(k(a1, . . . , an)/k(x1, . . . , xn)) = Sn. Therefore Sn occurs regularly over k. Hence if k is Hilbertian then Sn occurs as the Galois group over k.

Given that the Inverse Galois Problem asks about ﬁnding groups as Galois groups of extensions of Q, the importance of this next result speaks for itself. I quote from [12] without proof Theorem 3.3.4. (Hilbert’s Irreducibility Theorem) The ﬁeld Q is hilbertian.

As a corollary to this we see that Sn occurs as a Galois group over Q. In general, if we can realise a group over the rational function field Q(x) then Hilbert’s irreducibilty theorem coupled with Theorem 3.3.1 says that G occurs as a Galois group over Q. Now Riemann’s Existence Theorem will give a way of realising groups over the complex function field C(x). This motivates some discussion on decent of the base field, in other words, if G occurs as a Galois group over C(x), when can we say that G occurs as a Galois group over Q(x)? To answer this question one first needs to introduce rigidity.

As a digression, the chapter concludes with a nice little application of Hilbert’s Irreducibil- ity Theorem. Consider the polynomial f(x, y) = y2 − g(x). Theorem 3.3.4 says that if g(b) is a perfect square in Q for all b ∈ Q then g is the square of a polynomial [14].

20 Chapter 4

Rigidity

This chapter investigates the idea of branch points and how they are used in the development of an algebraic ramification type. This will finally allow the algebraic version of Riemann’s Existence Theorem to be stated. It is at this point that rigidity will be defined, and a first glimpse at the role it plays in the Inverse Galois Problem will be observed. The chapter ends with an important result that allows a character theoretic approach to be applied. This is used in many applications of rigidity in conjunction with [1].

4.1 Laurent Series Fields

Let k be a ﬁeld.

Deﬁne Λ the set of all sequences (ai)i∈Z with ai ∈ k such that ∃ N ∈ Z with ai = 0 for some i < N.

Then Λ is a ﬁeld with addition (ai)i∈ +(bi)i∈ = (ai+bi)i∈ and multiplication (ai)i∈ ·(bj)j∈ = P Z Z Z Z Z (cn)n∈Z where cn = i+j=n aibj [12] k is a subﬁeld of Λ. Indeed, can embed k in Λ by a ∈ k 7→ (ai)i∈Z with a0 = a and ai = 0 otherwise.

Deﬁne a sequence t by (ai) with a1 = 1 and ai = 0 else.

Deﬁne k[t] the ring generated by k and t. Then k[t] is a subring of Λ, it is the polynomial ring in one variable over k.

As a purely formal notation write, X i (ai) = ait The field operations of addition and multiplication defined correspond to formal addition and multiplication of Laurent Series and thus Λ is called the field of formal Laurent series over k and denoted by k((t)).

P i The ring of formal power series over k consisting of all ait is denoted k[[t]].

21 Polynomials over k[[t]] Have a ring homomorphism

ϕ : k[[t]] −→ k

(ai) 7→ a0

Consider a polynomial F (y) ∈ k[[t]][y]. By applying ϕ to all it’s coeﬃcients we obtain Fϕ(y) ∈ k[y]

Lemma 4.1.1. Let F (y) ∈ k[[t]][y] monic. If Fϕ(y) ∈ k[y] factors as

Fϕ = g · h with g, h ∈ k[y] and g, h coprime, then F factors as F = G · H with G, H ∈ k[[t]][y] monic and Gϕ = g, Hϕ = h.

4.2 Finite Extensions of Λ

−1 −1 For each e ∈ Z \{0}, let Ze = {i/e : i ∈ Z}. Then Ze forms an additive group isomorphic to Z.

−1 Z a subgroup of Ze of index e

Deﬁne Λe the set of sequences (aj)j∈Ze−1 with aj ∈ k. As before, Λe is a ﬁeld. In fact there is an isomorphism ∼ Λe = Λ

(aj)j∈Ze−1 7→ (bi)i∈Z with bi = ai/e. Under this isomorphism the element t ∈ Λ τ ∈ Λe where

τ = (aj)j∈Ze−1 where a1/e = 1, aj = 0 else.

We have Λ a subﬁeld of Λe consisting of all sequences (aj)j∈Ze−1 with aj ∈ k : aj = 0 ∀ j∈ / Z. th Lemma 4.2.1. Suppose k contains a primitive e root of unity ζe. Then Λe is Galois over Λ e of degree e and it has cyclic Galois group. Furthermore Λe = Λ(τ) with τ = t.

Proof. Consider the automorphism of Λe, ω : P b τ i → P (b ζi)τ i i∈Z i i∈Z i e It fixes elements of the form P b τ i with b = 0 unless ζi = 1 =⇒ b = 0 unless e divides i. i∈Z i i e i That is, the fixed field of ω is Λ.

Therefore, ω an automorphism of Λe with ﬁxed ﬁeld Λ so Λe is Galois over Λ with Ga- th lois group hωi. Further, we have that ω has order e (since ζe is an e root of unity) hence [Λe : Λ] = |Gal(Λe/Λ)| = |hωi| = e.

Finally, NTS that no non-trivial element of G = Gal(Λe/Λ) ﬁxes τ. That is, no non-trivial r r r element of hωi ﬁxes τ. Indeed ω(τ) = ζeτ, therefore ω (τ) = ζe τ = τ =⇒ e|r =⇒ ω is trivial element of G. Thus Λe = Λ(τ).

22 Now I’ll prove a useful result following [12]. Lemma 4.2.2. Let k be an algebraically closed ﬁeld with characteristic 0. Suppose F (y) ∈ k[[t]][y] monic, non-constant polynomial. Then F has a root in Λe for some e ∈ N.

Proof. Suppose F of minimal degree and has no root in any Λe, for a contradiction. Then n = deg F ≥ 2. Have n n−1 F (y) = y + λn−1y + ··· + λ1y + λ0 (λr ∈ k[[t]]) Can eliminate yn−1 term by replacing F with

λn−1 Fe(y) = F (y − ) n to assume F has zero yn−1 coeﬃcient.

n Claim If Fϕ(y) 6= y then F factors as F = GH with G, H ∈ k[[t]][y] monic, non-constant polynomials.

Indeed, since k is algebraically closed the polynomial Fϕ ∈ k[y] factors as a product of monic linear polynomials. If these factors are not all equal then can write Fϕ = g · h as in 4.1.1 and n thus the claim follows. However, if all the linear factors are the same, i.e. if Fϕ(y) = (y − a) n−1 n with a ∈ k, then the y coeﬃcient is −na hence a = 0 (as char k = 0) and Fϕ(y) = y contradicting the assumption.

n n So if Fϕ(y) 6= y then F factors as above, contradicting the minimality of F . Therefore Fϕ = y which is to say that ϕ(λr) = 0 ∀ r =⇒ all λr have zero constant term.

Now, it cannot be the case that all terms in all coefficients λr are zero as then would have n λr = 0 ∀ r and hence F (y) = y which has zero as a root contradicting our hypothesis. Con- sider only those λr 6= 0. Define mr the lowest power of t occurring with non-zero coefficient in λr, that is

mr λr = a · t + higher terms with a ∈ k\{0}. Deﬁne m u := inf r r n − r

Then u ∈ Q. Write u = d/e with d, e ∈ Z>0. Now consider the polynomial

n−2 ∗ −dn d n X d(r−n) r F (y) = τ F (τ y) = y + λrτ y ∈ Λe[y] r=0 The yr coeﬃcient is a Laurent series in τ of the form

d(r−n) mr d(r−n) λrτ = a · t τ + higher terms = a · τ Er + higher terms

mr where Er = e(n − r)( n−r − u). We have that Er = 0 for at least one r. Hence the coeﬃcients of F ∗ are power series in τ, and for at least one r, this power series has nonzero constant term. ∗ n ∗ Thus Fϕ 6= y , so by the above claim F = GH factors over k[[t]]. Then deg H < n = deg F 1/e0 ∗ 1/e0 hence has a root in Λe(τ ) by minimality of n. Thus F has a root in Λ(τ ) = Λee0 . Thus F also has a root in Λee0 .

23 Theorem 4.2.3. Let k an algebraically closed field, characteristic 0. Let ∆ be a field extension of Λ = k((t)) of finite degree e. Then ∆ = Λ(δ) with δe = t. Proof. Let γ be a primitive element for the finite extension ∆/Λ, that is write ∆ = Λ(γ). Then γ satisfies an irreducible polynomial F (y) ∈ Λ[y]. WLOG we may assume that F (y) is a monic, non-constant polynomial. Then by 4.2.2, F has a root γ1 in some Λe1 ∴ ∆ ⊂ Λe1 .

Now, Gal(Λe1 /Λ) is cyclic order e1, hence for each e diciding e1 there exists an intermediate e ﬁeld between Λ and Λe1 of degree e over Λ. Therefore ∆ = Λe = Λ(δ) where δ = t. Remark 4.2.4. The theorem says that for every ﬁnite extension ∆/Λ, it is the case that ∼ ∆ = Λe, for some e. But this is just what was proved in 4.2.2, that each polynomial over Λ has a root in some Λe.

4.3 Branch Points

th Deﬁnition 4.3.1. A compatible system (ζe)e∈N consists of primative e roots of unity such e2 that whenever e = e1 · e2 then ζe = ζe1 k Set P1 = k ∪ {∞}.

1 For p ∈ Pk deﬁne an isomorphism

vp : k(x) → k(t) a 7→ a ∀ a ∈ k t + p p 6= ∞ x 7→ 1/t p = ∞

1 Let L/k(x) be a finite Galois extension. Set G = Gal(L/k(x)), and let p ∈ Pk. Definition 4.3.2. Let γ be a primitive element of L/k(x). Then γ satisfies an irreducible polynomial F (y) ∈ k(x)[y]. Let Fvp be the polynomial obtained by applying vp to the coefficients of F (y).

Lemma 4.3.3. We can extend vp to an isomorphism v : L → Lv, where Lv is a subﬁeld of some FG-extension ∆ of Λ. Then the group Gal(∆/Λ) leaves Lv invariant.

Proof. Let H be an irreducible factor of Fvp , and deﬁne . ∆ := Λ[y] (H)

a ﬁnite extension of Λ. Now H has a root γ1 in ∆ and therefore so does Fvp . Therefore we can extend vp to an isomorphism v from L = k(x)[y] to Lv = k(t)[γ1] ⊂ ∆.

Now Lv is generated by the roots of Fvp , and since Gal(∆/Λ) permutes these roots then Gal(∆/Λ) leaves Lv invariant.

−1 Deﬁne gv := v · ω · v, where ω is the generator of Gal(∆/Λ). Then gv ∈ G. Lemma 4.3.4. If ∆ is another FG-extension of Λ with subﬁeld L , and v : L → L is an e ve e ve isomorphism extending v , then g and g lie in the same conjugacy class of G. p v ve

24 Proof. We may assume that ∆ and ∆ both lie in a FG-extension ∆ of Λ. Then L and L are e 0 v ve both subﬁelds of ∆ generated over k(t) by roots of F in ∆ . Hence L = L . 0 vp 0 v ve

Set h = v−1v. Then h ∈ G. Indeed, v : L → L = L and v−1 : L → L, isomorphisms. e e ve v v −1 Hence h = v · ve : L → L and thus h ∈ G

Let ω0 be generator of Gal(∆0/Λ). Restriction to ∆ (resp. ∆)e gives the generator of Gal(∆/Λ) (resp. Gal(∆e /Λ)). We have

g = v−1 · ω · v = h−1v−1 · ω · vh = h−1 · g · h ve e 0 e 0 v Therefore g and g lie in the same conjugacy class, say C . ve v p

Deﬁnition 4.3.5. Call the conjugacy class containing g and g in the proof of 4.3.4 the class v ve of G associated with p. Denote Cp. Once L is ﬁxed, this class depends only on p.

Deﬁnition 4.3.6. Let eL,p be the (common) order of elements of Cp. Call this the ramiﬁcation index of L at p. If L and p are clear from context then drop the subscript and just write e.

Lemma 4.3.7. All irreducible factors H of Fvp in Λ[y] have degree e. Furthermore, we can take ∆ as in 4.3.3 to be ∆ = Λe.

Proof. Can restrict Gal(∆/Λ) → Gal(Lv/k(t)). This is an injective homomorphism. Indeed if we write ∆ = Λ(γ1) we have Gal(Λ(γ1)/Λ) → Gal(k(t)[γ1]/k(t)) which has trivial kernel.

The degree [∆/Λ] =order of ω =order of ω|Lv .

−1 Now gv = v ωv hence order of ωLv = order of gv = e.

∼ Finally, by Theorem 4.2.3, since [∆/Λ] = e, we have ∆ = Λe.

1 Deﬁnition 4.3.8. Let L/k(x) be a FG-extension, and let p ∈ Pk. Then p is a branch point of L/k(x) if eL,p > 1. Remark 4.3.9. That is the same as saying that the class of G associated to p is non-trivial.

Lemma 4.3.10. Let L/k(x) and L0/k(x) be FG-extensions such that L0 ⊂ L. Let ρ : G → G0 by restriction, where G = Gal(L/k(x)), G0 = Gal(L0/k(x)). Then ρ maps the class of G associated 0 0 with p to the class of G associated with p. That is, ρ(Cp) = Cp. 0 0 Proof. Let v : L → ∆ as in Lemma 4.3.3. Then restrict to L to get v = v|L0 , an isomorphism 0 from L to a subset of ∆, extending vp.

0 −1 0 −1 0 Therefore gv0 = (v ) ωv = v ωv|L0 = (gv)|L0 . Therefore restriction to L does indeed map 0 Cp to Cp, and the lemma is proved. Theorem 4.3.11. Branch cycle argument 0 Let L/k(x) and L /k(x) be FG-extensions of degree n. Let α ∈ Aut(k) and let m ∈ Z such that −1 m 0 ∗ α (ζn) = ζn . Suppose α extends to an isomorphism λ : L → L ﬁxing x. Let λ denote the induced group isomorphism from G to G0 (in the usual notation) mapping g 7→ λgλ−1. Then,

0 ∗ m Cα(p) = λ (Cp)

25 1 Proof. Fix p ∈ Pk. Then e = eL,p divides n = |G| by Lagrange, say n = e · s. Then by compatibility of ζi’s we have

s −1 −1 s ms m ζn = ζe ⇒ α (ζe) = α (ζn) = ζn = ζe (4.1)

i i Extend α to an automorphism of Λe = k((τ)) sending Σbiτ 7→ Σα(bi)τ . Call this automorphism αe. As before let ω be the generator of G = Gal(Λe/Λ) with ω(τ) = ζeτ. Then we have, using (4.1) −1 −1 −1 m m αe ωαe(τ) = αe ω(τ) = αe (ζeτ) = ζe τ = ω (τ) 0 −1 By 4.3.7 can extend vp to an isomorphism v from L to a subfield of Λe. Then define v := αe·v·λ , 0 an isomorphism extending vα(p) from L to a subfield of Λe. Indeed it fixes all K ∈ k since λ is −1 −1 an extension of α and hence λ (K) = α (K) and αe sends K 7→ α(K).

Consider,

0 −1 v (x) = αvλe (x) = αve (x) (since λ is identity on x) α(t + p) p 6= ∞ = e αe(1/t) p = ∞ t + α(p) p 6= ∞ = 1/t p = ∞

0 Now gv0 ∈ Cα(p) as before is

0 −1 0 −1 −1 −1 gv0 = (v ) ωv = λv αe · ω · αvλe = λv−1 · ωm · vλ−1 m −1 = λ · gv λ ∗ m = λ (gv) proving the theorem.

4.4 Rigid Ramiﬁcation Types

1 Let k be an algebraically closed subset of C and let P = C ∪ {∞}

1 Consider the triple (G, P, C) where G is a group, P is a ﬁnite subset of P and C = (Cp)p∈P . Deﬁne an equivalence relation on this triple by (G, P, C) ∼ (G0,P 0, C0)

0 0 0 iff ∃ an isomorphism from G → G which sends Cp 7→ Cp for each p ∈ P and P = P . (This is clearly an equivalence relation!) Definition 4.4.1. The set of equivalence classes of (G, P, C) is called a (algebraic) ramification type and denoted T = [G, P, C].

1 Remark 4.4.2. For a FG-extension L/k(x) and a set P ⊂ P we have canonical choice for a triple as above; we take G = Gal(L/k(x)) and C the set of classes of G associated with p, the 1 branch points of L/k(x) over P . Therefore, the Inverse Galois Problem can be reformulated in this language asking if one can ﬁnd an extension of Q of type T = [G, P, Cp] for any group G.

26 Recall the topological version of Riemann’s Existence Theorem 2.3.8. There is an algebraic analogue in which the topological ramification type and the algebraic ramification type are identified using Riemann surface theory. Under this identification there is an algebraic version of Riemann’s Existence Theorem [12]. Theorem 4.4.3. Riemann’s Existence Theorem Let T = [G, P, C] be a type and say P = {p1, . . . , pr}. Then there exists an FG-extension of

C(x) of type T iff ∃ generators σ1, . . . , σr of G with σ1 ··· σr = 1 and σi ∈ Cpi for i = 1, . . . , r. So if one can find a generating set of a finite group G satisfying these properties then Riemann’s Existence Theorem guarantees the existence of an extension of C(x) of the desired type. I now introduce the notion of rigidity as an extra condition on this generating set. As a first application, I will show that rigidity ensures a unique extension of desired type in Theorem 4.4.3, but as we will see in later chapters rigidity will be crucial in guaranteeing an extension of Q of the desired type, which is of course the aim.

Deﬁnition 4.4.4. Let (C1,...,Cr) be a tuple of conjugacy classes in group G. We say it is a rigid tuple if

1. There exist generators σ1, . . . , σr of G with σ1 ··· σr = 1 and σi ∈ Ci for i = 1, . . . , r.

0 0 2. If σ1, . . . , σr is another set of generators with that property then there exists a unique −1 0 g ∈ G such that gσig = σi. The uniqueness of g is equivalent to saying G has trivial center. Indeed if 1 6= z ∈ Z(G) and −1 0 −1 −1 −1 −1 0 g ∈ G as in deﬁnition, then gσig = σ1 ⇒ (zg)σi(zg) = z(gσig )z = zσiz = σi, since 0 z commutes with σi. But then g is not unique, which is a contradiction. −1 0 Deﬁnition 4.4.5. If instead of a unique element g ∈ G such that gσig = σi, there exists a 0 general automorphism γ ∈ Aut(G) such that γ(σi) = σi, then we say (C1,...,Cr) is a weakly rigid tuple.

Since γ ∈ Aut(G) is deﬁned on (σ1, . . . , σr), and these generate G, then γ is uniquely n n deﬁned on all G. Indeed, let h ∈ G. Then since {σ } generate G can write h = σ i1 ··· σ ik , i i1 ik with ij ∈ [1, . . . , r] and nij ∈ Z, so

ni ni n n γ(h) = γ(σ 1 ··· σ k ) = γ(σ ) i1 ··· γ(σ ) ik i1 ik i1 ik

Lemma 4.4.6. Let (C1,...,Cr) be a rigid tuple of conjugacy classes, and let γ ∈ Aut(G) such that γ fixes each Ci. Then γ ∈ Inn(G) is an inner automorphism, i.e. acts by conjugation. 0 0 0 0 Proof. Define σi = γ(σi). Then σi ∈ Ci since γ fixes each Ci, and σ1 ··· σr = γ(σ1) ··· γ(σr) = 0 0 γ(σ1 ··· σr) = γ(1) = 1. Let h ∈ G, then ∃ h ∈ G such that h = γ(h ) since γ is an automor- n n phism. Now, h0 = σ i1 ··· σ ik , since G is generated by the σ , so we have i1 ik i

0 ni ni 0 n 0 n h = γ(h ) = γ(σ 1 ··· σ k ) = (σ ) i1 ··· (σ ) ik i1 ik i1 ik 0 So the σ generate G too. Now by rigidity of (C1,...,Cr) we have that ∃ g ∈ G such that −1 0 −1 gσig = σi. But by remarks made above, since γ is unique, for x ∈ G, γ(x) = gxg . That is, γ is an inner automorphism.

Deﬁnition 4.4.7. A type T = [G, P, C] is called rigid if the classes Ci form a rigid tuple.

Theorem 4.4.8. For each rigid type there is a unique FG-extension of C(x) of this type up to C(x) isomorphism.

27 Proof. Existence follows from RET. For uniqueness, suppose L1/C(x) and L2/C(x) are both of the same type, and suppose that type is rigid. WLOG let L1 and L2 lie in a common FG- extension L of C(x). Let G = Gal(L/C(x)), Gi = Gal(Li/C(x)) for i = 1, 2. Then deﬁne (i) ρi : G → Gi by restriction, and for p ∈ P, denote by Cp (resp. Cp ) the class of G (resp. Gi) (i) associated with p. Then ρi(Cp) = Cp by 4.3.10. Let p1, . . . , pr be the brach points of L.

By RET there exist generators σ1, . . . , σr of G with the properties that σ1 ··· σr = 1 and

σj ∈ Cpj . Then ρi(σ1), . . . , ρi(σr) generate Gi. It is clear that they also satisfy the same properties.

L1 and L2 are of the same type, so by deﬁnition there exists an isomorphism : G2 → G1 (2) (1) that maps Cp 7→ Cp .

Consider (ρ2(σ1)), . . . , (ρ2(σr)). They clearly generate G1. We have

(ρ2(σ1)) ··· (ρ2(σr)) = (ρ2(σ1) ··· ρ2(σr))

= (ρ2(σ1 ··· σr))

= (ρ2(1)) = 1

(2) (1) Also, σi ∈ Cpi ⇒ ρ2(σi) ∈ Cpi ⇒ (ρ2(σi)) ∈ Cpi .

So (ρ2(σ1)), . . . , (ρ2(σr)) satisfy the same properties as ρ1(σ1), . . . , ρ1(σr). So by deﬁnition −1 4.4.4 (of rigidity) ∃ g ∈ G1 : g · (ρ2(σi)) · g = ρ1(σi) for i = 1, . . . , r. Deﬁne,

δ : G1 → G1 σ 7→ σg = g · σ · g−1 action by conjugation. So we have δ((ρ2(σi))) = ρ1(σi) Then we have an isomorphism γ := δ · : G2 → G1 such that

γ(ρ2(σi)) = δ · (ρ2(σi))

= ρ1(σi)

That is ρ1 = γ · ρ2, so ρ1 and ρ2 have the same kernel. The ﬁxed ﬁeld in L of this kernel equals L1 and L2, therefore L1 = L2. Remark 4.4.9. We could actually prove a stronger theorem by replacing the condition of rigidity with weakly rigid. The proof goes the same way except δ is just a general automorphism.

4.5 Structure Constants

It can in general be difficult to find rigid tuples by hand. In this section I present a formula that translates the condition of uniqueness of generators up to conjugacy (i.e. Definition 4.4.4(2)) into a condition on the characters of the conjugacy classes. This can be checked using the character tables of groups in [1] or using a computer (e.g. using GAP). I conclude the chapter by using the aforementioned formula to find rigid tuples in SL2(8) and S4.

A class vector C is just a collection of conjugacy classes {C1,...,Cr}. Write σ = {σ1, . . . , σr}. So to say σ1, . . . , σr generate G is the same as saying hσi = G. Lets redeﬁne a rigid tuple as in [7]. Firstly,

28 Definition 4.5.1. Denote the set of all generating s-systems by Σs(G) i.e. Σs(G) := {σ ∈ s G : hσi = G, σ1 ··· σs = 1} Then define, Σ(C) := {σ ∈ Σs(G): σi ∈ Ci} So Σ(C) consists of generating s-systems that satisfy the first part of the definition of rigidity Definition 4.4.4(1). The second part of this definition is that if another such system σ0 exists then there exists an element g such that the action by conjugation by g on σ gives σ0. This action is an inner automorphism, and so motivates the following definition, . Σ(C) l(C) := Inn(G) Definition 4.5.2. A class vector C is called rigid if l(C) = 1 Now consider the set of not necessarily generating s-systems s Σ(b C) := {σ ∈ G : σi ∈ Ci, σ1 ··· σr = 1} g g G acts by component-wise conjugation, i.e. (σ1, . . . , σr) → ( σ1,..., σr), so similarly quotient out by Inn(G) and define, . Σ(b C) n(C) := Inn(G) Definition 4.5.3. n(C) is the normalised structure constant of C. Remark 4.5.4. Σ(b C) is an enlargement of Σ(C), so l(C) ≤ n(C) with equality only when Σ(b C) = Σ(C). Let . Qn := Σ(b C) Inn(G)

Proposition 4.5.5. If C is a class vector of conjugacy classes Ci of a ﬁnite group G then X Z(G) n(C) = CG(hσi) [σ]∈Qn

Proof. Let G act on Σ(b C) by conjugation (as above). The class equation gives, X |Σ(b C)| = |G : CG(hσi)| (4.2)

[σ]∈Qn Now,

|Σ(b C)| n(C) = |Inn(G)| X |G : CG(hσi)| = (by 4.2) |Inn(G)| [σ]∈Qn X |G|/|Inn(G)| = |CG(hσi)| [σ]∈Qn . Now G Z(G) =∼ Inn(G) ⇒ |G|/|Inn(G)| = |Z(G)| ⇒ result.1

1 This follows from the ﬁrst isomorphism theorem, let h : G → Aut(G) by g 7→ πg then Im(h) = Inn(G) and Ker(h) = Z(G).

29 Theorem 4.5.6. Let C a class vector in a ﬁnite group G, and let s ≥ 2. Then

s−2 s X |G| Y χ(σi) n(C) = |Z(G)| s−2 (σi ∈ Ci) χ(1) |CG(σi)| χ∈Irr(G) i=1

Proof. χ ∈ Irr(G). Let R : G → GLn(C) be corresponding representation. Schur’s lemma ⇒ for each σ ∈ G ∃ ω(σ) ∈ C such that 1 X χ(σ) R(σρ) = ω(σ)I (ω(σ) = ) |G| n χ(1) ρ∈G It is clear then that for σ, τ ∈ G

1 X χ(σ) R(σρτ) = ω(σ)I · R(τ) = R(τ) |G| n χ(1) ρ∈G

2 Furthermore, for ρ = (ρ1, ρ2) ∈ G , σ1, σ2, τ ∈ G we have,

1 X χ(σ1) χ(σ2) R(σρ1 σρ2 τ) = · · R(τ) |G|2 1 2 χ(1) χ(1) ρ∈G2

By induction we can extend this, for ρ = (ρ1, . . . , ρs), to,

1 X χ(σ1) ··· χ(σs) R(σρ1 , . . . , σρs τ) = R(τ) |G|s 1 s χ(1)s ρ∈Gs Set τ = 1 and take traces,

1 X χ(σ1) ··· χ(σs) χ(σ1) ··· χ(σs) χ(σρ1 , . . . , σρs ) = χ(1) = |G|s 1 s χ(1)s χ(1)s−1 ρ∈Gs

Multiply both sides by χ(1)|G|s−1. LHS becomes

1 X 1 X χ(1)|G|s−1 χ(σρ1 ··· σρs ) = χ(1)χ(σρ1 ··· σρs ) |G|s 1 s |G| 1 s ρ∈Gs ρ∈Gs Hence equation becomes,

1 X χ(σ1) ··· χ(σs) χ(1)χ(σρ1 ··· σρs ) = |G|s−1 (4.3) |G| 1 s χ(1)s−2 ρ∈Gs Deﬁne X 1 X m(C) := χ(1)χ(σρ1 ··· σρs ) |G| 1 s ρ∈Gs χ∈Irr(G) Summing over both sides of (4.3) over χ ∈ Irr(G) gives

X χ(σ1) ··· χ(σs) m(C) = |G|s−1 χ(1)s−2 χ∈Irr(G)

Now 1 n(C) = |{σ ∈ C : σ ··· σ = 1}| |Inn(G)| 1 s

30 may be written as s m(C) Y 1 n(C) = |Inn(G)| |C (σ )| i=1 G i |G| 1 |Z(G)| And so, recalling that = |Z(G)| =⇒ = we get, |Irr(G)| |Irr(G)| |G|

s |Z(G)| Y 1 n(C) = m(C) |G| |C (σ )| i=1 G i |Z(G)| X χ(σ1) ··· χ(σs) = |G|s−1 |G| χ(1)s−2 χ∈Irr(G) s−2 s X |G| Y χ(σi) = |Z(G)| s−2 χ(1) |CG(σi)| χ∈Irr(G) i=1

And the theorem is proved.

This next result is key and is used frequently in applications of rigidity [7].

Corollary 4.5.7. A class vector is C is rigid if the following hold:

1. G = hσi with σi ∈ Ci : σ1 ··· σs = 1

X χ(σ1) ··· χ(σs) |CG(σ1)| · · · |CG(σs)| 2. = χ(1)s−2 |G|s−2|Z(G)| χ∈Irr(G) Proof. (2.) ⇒ n(C) = 1. Recall from remark 4.5.4 that l(C) ≤ n(C). Well (1.) ⇒ l(C) ≥ 1. Therefore, together we get that l(C) = 1 and so C is a rigid class vector.

Examples

1. The Special Linear Group, SL2(8) Using GAP to compute the character table of SL2(8) (see Appendix A, Figure A.1) let’s verify that the class vector C = (9a, 9b, 9c) is rigid; ﬁrstly lets check condition (2.):

1 · 1 · 1 1 · 1 · 1 ABC BCA CAB −1 · −1 · −1 LHS = + + + + + 1 7 7 7 7 8 1 1 1 1 1 81 = (1 + + + + − ) = 7 7 7 7 8 56 Now to calculate the sizes of the centralisers, the size of the conjugacy classes have been found, again in Appendix A, Figure A.1, together with the formula |Cl(σ)| = |G : CG(σ)|. Hence 9 · 9 · 9 729 81 RHS = = = 504 · 1 504 56 and thus condition (2.) is satisﬁed. Now lets verify condition (1.):

Assume for a contradiction that the condition is false, that is, σ ∈ C : σ1σ2σ3 = 1 ⇒ hσi ( G. Only maximal subgroups of G with order divisible by 9 are isomorphic to D18 2 7 4 5 [7] and hence hσi = Z9 ⇒ σ2 either σ1 or σ1 and σ3 either σ1 or σ1.

31 i j But σ1σ2σ3 = σ1σ1σ1 6= 1 for i=2 or 7, j=4 or 5 ⇒⇐ Contradiction.

And so we have veriﬁed that C is a rigid class vector in SL2(8). Now we have a rigid type with G = SL2(8) and thus there exists a unique FG-extension of C(x) of this type upto C(x)-isomorphism.

2. The Symmetric Group, S4 (2) (3) (4) Let σ1 = (34), σ2 = (123), σ3 = (3421). Lets verify that C = (C ,C ,C ) forms a rigid class vector.

First show that hσi = S4. One calculates

σ2 σ2 σ3 σ1 = (1 4) (1 4) = (2 4) (2 4) = (1 2) (1 2)σ3 = (1 3) (1 2)σ2 = (2 3) (2 3)σ3 = (1 4)

and hence by conjugation one sees that all transpositions are in hσi and therefore hσi = S4. Moreover σ1σ2σ3 = 1 by inspection. It remains to verify the second condition using the character table of S4 which I generate using GAP, and using the formula Corollary 4.5.7

X χ(σ1) ··· χ(σs) |CG(σ1)| · · · |CG(σs)| = χ(1)s−2 |G|s−2|Z(G)| χ∈Irr(G)

See Appendix A, Figure A.2 for the character table of S4, which we can use to calculate, X χ(σ1)χ(σ2)χ(σ3) 1 · 1 · 1 (−1) · 1 · (−1) LHS = = + = 2 χ(1)3−2 1 1 χ∈Irr(G)

and for the size of the conjugacy classes, which can be used to calculate,

|CG((3 4))| = 4, |CG((1 2 3))| = 3, |CG((3 4 2 1))| = 4.

Therefore,

|C ((3 4))||C ((1 2 3))||C ((3 4 2 1))| 4 · 3 · 4 RHS = G G G = = 2 |S4||Z(S4)| 24 · 1

and thus Corollary 4.5.7 gives that C is a rigid class vector in Sn

32 Chapter 5

The Rigidity Criteria

Now we have deﬁned rigidity, and found some rigid class vectors. Riemann’s Existence Theo- rem then guarantees the existence of an extension over C(x) of the desired type, but this hasn’t told us anything about extensions of Q. In this chapter the importance of rigidity in our context of trying to ﬁnd groups as Galois groups over Q is revealed in the Rational Rigidity Criteria.

Riemann’s Existence Theorem and Hilbert’s Irreducibility Theorem are combined to give a condition on G, a finite group, that if satisfied gives a positive solution to the Inverse Galois Problem for G. If G forms a rigid type, Riemann’s Existence Theorem gives an extension over C(x) of that type. In this chapter a condition, rational rigidity, is shown to ensure an extension of Q(x) of that same type. Hilbert’s Irreducibility Theorem provides the final step by concluding that, since Q is hilbertian, G is realisable over Q.

5.1 Descent of the Base Field

Definition 5.1.1. L/k(x) is defined over κ if there exists subfield Lκ ⊆ L, Galois over κ(x) and regular over κ and such that [L : k(x)] = [Lκ : κ(x)]

Suppose L is defined over κ. Consider the Galois extension Lκ/κ(x). Let θ be a primitive element of the extension, that is Lκ = κ(x)(θ). That is to say, θ satisfies an irreducible polynomial F (y) ∈ κ(x)[y] of degree n = [L : k(x)]. By Lemma 3.1.4, since Lκ is regular over κ by definition, F (y) remains irreducible over κ0(x)[y], where κ0 is an intermediate field between 0 κ and k. Now the roots of F (y) are all contained in Lκ0 , so Lκ0 is Galois over κ (x). The 0 0 irreducibility of F (y) over κ (x) implies that Lκ is a degree n extension over κ (x), and it is geometric too by Lemma 3.1.4. Therefore L is defined over κ0, for any intermediate field κ0 between κ and k. This proves the following result:

Lemma 5.1.2. If L is defined over κ and θ is a primitive element of the extension Lκ/κ(x) then L = k(x)(θ) Proof. Simply notice that k can be seen as an intermediate field κ0, thus L is defined over k. Hence k(x)(θ) is an extension of degree n, but it is a subfield of L, therefore L = k(x)(θ) as required.

So we have that if L is defined over κ then it is defined over all intermediate fields κ ⊆ κ0 ⊆ k, and the field Lκ0 was generated by the roots of the minimal polynomial F (y) of the primitive element θ for the extension Lκ/κ(x). But by Lemma 5.1.2 L = k(x)(θ) and so the Galois group Gal(L/k(x)) acts naturally to permute these roots. The following result is immediately obtained

33 ∼ Lemma 5.1.3. Gal(L/k(x)) = Gal(Lκ/κ(x)) via restriction to Lκ. And so one sees the utility of so called fields of definition in solving the Inverse Galois problem as stated, to find groups as Galois groups over Q, given that Riemann’s Existence Theorem only guarantees extensions of C(x) of desired type. Lemma 5.1.4. Let L be defined over κ and suppose that either Gal(L/k(x)) has trivial center or that κ is algebraically closed. Then Lκ is the unique subfield of L, Galois over κ(x) of degree n and regular over κ.

Proof. Suppose that Leκ is another subfield of L satisfying the properties of Definition 5.1.1. Consider the subfield K of L generated by Leκ and Lκ. Since these are both Galois over κ(x) then so is K . Let κ0 be the algebraic closure of κ in K. Then κ0 and thus κ0(x) are invariant 0 under Gal(K/κ(x)), so κ (x) is Galois over κ(x). Let θ be a primitive element for Lκ/κ(x) as 0 0 in Lemma 5.1.2. Then, Lκ ⊆ K ⇒ |K : κ(x)| ≥ |κ (x)(θ): κ (x)| = n. Let γ be a primitive element for K/κ0(x). By Lemma 3.1.4 the minimal polynomial for γ remains irreducible over k(x). Therefore |K : κ0(x)| ≤ n = |L/k(x)|. Therefore |K/κ0(x)| = n. That is K = κ0(x)(θ).

0 0 If κ is algebraically closed then κ = κ. So K = κ (x)(θ) = κ(x)(θ) = Lκ by deﬁnition of θ. So indeed Lκ is unique.

0 0 Now suppose Gal(L/k(x)) has trivial center. We have that κ (x) ∩ Lκ = κ (x) ∩ κ(x)(θ) = κ(x) since κ is algebraically closed in Lκ by deﬁnition. The same holds for Leκ by assumption that Leκ satisﬁes the same properties as Lκ. Hence by the Galois correspondence

0 Gal(K/κ(x)) = Gal(K/Lκ) × Gal(K/κ (x)) 0 Gal(K/κ(x)) = Gal(K/Leκ) × Gal(K/κ (x))

Moreover, by Lemma 5.1.3, Gal(K/κ0(x)) =∼ Gal(L/k(x)), which by assumption has trivial 0 center. Therefore Gal(K/Lκ) = G(K/Leκ) = CGal(K/κ(x))(Gal(K/κ (x))). Therefore Lκ = Leκ and indeed Lκ is unique. Consider the finite Galois extension L/k(x) with primitive element θ satisfying the irre- n ducible polynomial F (y) ∈ k(x)[y]. Let {θi}i=1 be the roots of F . There exists polynomials fi(y) ∈ k(x)[y] such that θi = fi(θ). The coefficients of F and the fi lie in k(x) and so lie in 0 0 0 0 κ (x) for a finitely generated extension κ of κ. Let Lκ0 = κ (x)(θ), Galois over κ (x) and regular over κ0. Hence L is always defined over a finitely generated extension κ0 of κ.

Lemma 5.1.5. Suppose M/κ(x) is a finite Galois extension which is regular over κ. Then there ∼ exists a finite Galois extension L/k(x), defined over κ with Lκ = M, where Lκ as in Definition 5.1.1.

Proof. Let γ be a primitive element for the extension M/κ(x). Define L := k(x)(γ), a finite Galois extension of k(x) of the same degree as M/κ(x). Now by the same argument as in Lemma 5.1.2 in reverse, get that L is defined over κ with Lκ = M.

1 Have P = C ∪ {∞}. Recall that to a finite Galois extension L/k(x) we associate a triple (G, P, C) where G = Gal(L/k(x)), P is the set of branch points and Cp is the class of G associated to p ∈ P . It is from this association that the notion of ramification type arises. Now how does the ramification type of L/k(x) relate to that of Lκ/κ(x) when L is defined over κ? If κ is algebraically closed the following theorem answers this,

34 Theorem 5.1.6.

1. Let L be deﬁned over κ. Then L/k(x) is naturally of the same ramiﬁcation type at Lκ/κ(x) if κ is algebraically closed.

2. If K is an algebraically closed subﬁeld of C then for each weakly rigid type T there is at most one extension of K(x) of this type.

Proof.

1. Let θ be a primitive element for Lκ/κ(x). Then by Lemma 5.1.2, since L is defined over κ, L = k(x)(θ). Let p ∈ κ ∪ {∞} and define vp : k(x) → k(t) as when defining branch points. Similrly, this extends to v : L → ∆, where ∆ is a finite Galois extension of Λ = k((t)). Now v(κ(x)) = κ(t) and hence v(Lκ) = v(κ(x)(θ)) = κ(t)(v(θ)). Thus, by restricting to Lκ one obtains ve : Lκ → ∆κ where ∆κ := κ((t))(v(θ)). Hence restriction yields an isomorphism between Gal(∆/Λ) and Gal(∆κ/κ((t))), mapping the class Cp of Gal(L/k(x)) associated with p to the class Cep of Gal(Lκ/κ(x)) associated with p, for all p ∈ κ∪{∞}. So the question now is, where are the branch points of L/k(x) and Lκ/κ(x)? θ satisfies monic irreducible polynomial F (y) ∈ κ(x)[y]. The discriminant D(x) ∈ κ(x) is non-zero because F is irreducible and hence separable. Let p ∈ κ with D(p) 6= 0. View F (y) ∈ κ(x)[y] as F (x, y) ∈ κ[x, y]. Then (vpF )(y) = F (t + p, y). So (vpF ) is a monic polynomial in y with coefficients in κ[t]. By Lemma 4.1.1 then vpF factors as a product of linear factors in Λ[y] if F (p, y) is separable, which is the case when D(p) 6= 0. And these are precisely the points with eL,p = 1 by 4.3.7. Therefore, branch points, which are the points where eL,p 6= 1 are roots of D(x), and so are algebraic over κ. But κ is algebraically closed by assumption, so the branch points of L/k(x) lie in κ ∪ {∞}, and clearly so do the branch points of Lκ/κ(x). Thus the restriction isomorphism yields an equivalence of the triple (Gal(L/k(x)),P,Cp) with (Gal(Lκ/κ(x)),P, Cep) in other words both extensions are of the same ramification type.

2. Suppose for a contradiction that M and N are different finite Galois extensions of K(x) of weakly rigid ramification type T . By Lemma 5.1.5 there is a finite Galois extension L of C(x) defined over K such that LK is isomorphic to M, and similarly there exists 0 0 0 finite Galois extension L /C(x) such that LK is isomorphic to N. Then by (1) L and L 0 are of the same type, T . By Remark 4.4.9 then L and L are C(x)-isomorphic. Now by Lemma 5.1.4, since K is algebraically closed in C, we have that LK ⊂ L is unique, and 0 0 0 similarly, LK ⊂ L is unique. But L isomorphic to L implies that LK is isomorphic to 0 LK i.e. M = N, which is a contradiction.

α-Isomorphisms Consider α ∈ Aut(k). We can extend this naturally to an automorphism of k(x) fixing x. Suppose we have two finite Galois extenions of the same base field i.e.

L/k(x) L0/k(x) and suppose there is a field isomorphism λ : L → L0 which acts in the same way on the base fields of these extensions, as automorphisms of k(x), that is λ|k(x) = α. Such a field isomorphism

35 is called an α-isomorphism. By their very nature, each α-automorphism λ induces a group isomorphism

λ∗ : Gal(L/k(x)) → Gal(L0/k(x)) g 7→ λgλ−1

It is plain to see that this is indeed an isomorphism. It is a homomorphism since λ∗(g)λ∗(h) = λgλ−1λhλ−1 = λghλ−1 = λ∗(gh). Now λ : L → L0 is by deﬁnition a ﬁeld isomorphism, from which λ∗ is an isomorphism follows.

Let α ∈ Aut(k) and L/k(x) a ﬁnite Galois extension. Construct an α-isomorphism from α . as follows: Take L of the form k(x)[y] (F ) with F (y) ∈ k(x)[y] an irreducible polynomial. Extend α to an automorphism of k(x, y) ﬁxing x and y, denote this by αe. Now αe induces a ring isomorphism λ, . . λ : k(x)[y] → k(x)[y] (F ) (αFe )

Moreover λ|k(x) = α, so indeed λ is an α-isomorphism.

Lemma 5.1.7. Let L be deﬁned over κ and α ∈ Aut(k). If α|κ = Id then there exists an α-isomorphism λ : L → L with the induced map λ∗ = Id on Gal(L/k(x)).

Proof. L is deﬁned over κ so by 5.1.2, if θ is a primitive element of Lκ/κ(x) then L = k(x)(θ). That is, we have . . L = k(x)[y] (F ) Lκ = κ(x)[y] (F ) where F (y) ∈ κ(x)[y] is minimal polynomial of θ. Now construct an α-isomorphism as described above, where λ : L → L0 and . L0 = k(x)[y] (αFe ) 0 but by assumption α|κ = Id, so αe(F ) = F . Therefore L = L. Finally, for g ∈ Gal(L/k(x)), ∗ −1 have λ (g)(θ) = λgλ (θ). But λ ﬁxes Lκ, and g(θ) ∈ Lκ. So

λg(λ−1θ) = λ(gθ) = g(θ) so indeed λ∗ = Id.

Let G a group with N C G. If K ≤ G such that G = KN,K ∩ N = {1} and K C G then G = K × N. This is called a normal compliment [9].

Lemma 5.1.8. Let G be a group with N,K C G forming a normal compliment. Then G is a direct product G = K × N.

Proof. Consider [k, n] = k−1n−1kn

−1 −1 k n k ∈ N since N C G, so [k, n] ∈ N −1 n kn ∈ K since K C G, so [k, n] ∈ K

So [k, n] ∈ K ∩ N = {1}. This holds for all k ∈ K, n ∈ N so have [K,N] = 1. Now, since G = KN all g ∈ G can be written as g = kn for some k ∈ K, n ∈ N. I claim this expression is unique. Indeed suppose g = k1n1 = k2n2 for k1, k2 ∈ K, n1, n2 ∈ N. Then

36 −1 −1 −1 −1 k2 k1 = n2n1 ∈ K ∩ N = {1}. So k2 k1 = 1 ⇒ k1 = k2 and n2n1 = 1 ⇒ n1 = n2 so indeed the expression is unique. Deﬁne a map

ϕ : G → K × N g = kn 7→ (k, n)

Claim ϕ is an isomorphism: g1, g2 ∈ G then write g1 = k1n1, g2 = k2n2. Then g1g2 = k1n1k2n2 = k1k2n1n2 since [K,N] = 1. So ϕ(g1)ϕ(g2) = (k1, n1)(k2, n2) = (k1k2, n1n2) = ϕ(g1g2), so ϕ is a homomorphism. Since every g ∈ G is of the form kn for k ∈ K, n ∈ N ϕ is onto, hence is an isomorphism. This completes the proof.

Now we consider the descent from κ to κ, where κ is the algebraic closure of κ in C. As a technicality note that the group Gal(κ/κ) is naturally a proﬁnite group. For my purposes I view it as an abstract group, but it has the property that every for ﬁnite Galois extension κ0 of κ inside κ the restriction map Gal(κ/κ) → Gal(κ0/κ) is a surjection [12].

Lemma 5.1.9. Let k = κ and let L/k(x) be a ﬁnite Galois extension whose Galois group Gak(L/k(x)) has trivial center. Then L is deﬁned over κ if and only if for each α ∈ Gal(κ/κ) ∃ an α-isomorphism λ : L → L with λ∗ = Id.

Proof. Suppose L is defined over κ. α ∈ Gal(κ/κ) ⇒ α|κ = Id by definition of Galois group. Then by Lemma 5.1.7 there does indeed exists an α-isomorphism λ : L → L which induces a trivial map on the Galois group. Conversely, suppose for each α ∈ Gal(κ/κ) ∃ α-isomorphism λ : L → L with λ∗ = Id. L is always defined over a finitely generated extension κ1 of κ. Set L1 := Lκ1 . Let α1 ∈ Gal(κ1/κ). Since restriction Gal(κ/κ) is surjective we can extend α1 to an element α ∈ Gal(κ/κ), which by assumption has an α-isomorphism λ : L → L associated to it, such that λ∗ = Id. Gal(L/k(x)) has trivial center by hypothesis. Consider λ(L1). This is a subgroup of L, Galois over κ1(x) and regular over κ1. But Lemma 5.1.4 says that if L1 is the unique group with these properties.

Thus L1 = λ(L1). So λ1 := λ|L1 : L1 → L1. That is λ1 ∈ Aut(L1). But λ1 is an α1- ∼ isomorphism for α1 ∈ Gal(κ1/κ) = Gal(κ1(x)/κ(x)). Therefore (λ1)|κ(x) = α1 which ﬁxes ∗ −1 κ(x) and so λ1 ∈ Gal(L1/κ(x)). Now, λ = Id so λ1gλ1 = g for all g ∈ Gal(L/k(x)). But ∼ Gal(L/k(x)) = Gal(L1/κ1(x)) by 5.1.3. So for g1 ∈ Gal(L1/κ1(x)),

λ1g1 = g1λ1

Hence λ1 ∈ CGal(L /κ(x))(Gal(L1/κ1(x))) =: C. This holds for each α ∈ Gal(κ1/κ), so C 1 . Gal(L1/κ(x)) surjects onto Gal(L1/κ1(x)) := H/G, that is H = GC. Moreover G ∩ C lies in ∼ the center of Gal(L1/κ1(x)) = Gal(L/k(x)) which has trivial center, so G ∩ C = {1}. This is a normal compliment and so, by Lemma 5.1.8, we get that H = G × C, or in full,

Gal(L1/κ(x)) = Gal(L1, κ1(x)) × C ∼ Let Lκ be the fixed field of C in L1. Then Gal(Lκ/κ(x)) = H/C = G = Gal(L/k(x)). Now, Lκ is the fixed field of C by definition and κ1(x) is the fixed field of Gal(L1/κ1(x)) by definition of Galois group. Hence their intersection is the fixed field of CGal(L1/κ1(x)) = Gal(L1/κ(x)) which is κ(x). But Lκ ⊂ L1 and L1 is regular over κ1 therefore Lκ is regular over κ, showing that L is defined over κ, as required.

37 5.2 Rigidity Criteria

A standard definition in group theory is that of a rational conjugacy class; we say C, a conjugacy m class in a group G is called rational if C = C for all m ∈ Z coprime to the order of G [9]. Now we extend this definition to ramification types T ,

Definition 5.2.1. A ramification type T = [G, P, (Cp)p∈P ] is κ-rational if P ⊂ κ ∪ {∞} and m for each p ∈ P and α ∈ Gal(κ/κ), α(p) ∈ P and Cα(p) = Cp where m ∈ Z is such that −1 m th α (ζn) = ζn for ζn an n root of unity in κ where n = |G|. Notice the similarity between the condition in this definition and the condition in the Branch Cycle Argument, Theorem 4.3.11. The significance of this similarity is revealed in the next theorem,

Theorem 5.2.2. Let k = κ and let L/k(x) be an FG-extension with ramiﬁcation type T

1. If L is defined over κ then T defines a κ-rational ramification type.

2. If the ramiﬁcation type T is rigid and κ-rational, then L is deﬁned over κ.

Proof.

1. L is deﬁned over κ. Let α ∈ Gal(κ/κ). Then α|κ(x) = Id, so by Lemma 5.1.7 there is an ∗ −1 m α-isomorphism λ : L → L with λ = Id. Now let m ∈ Z such that α (ζn) = ζn . Then by the Theorem 4.3.11 (Branch Cycle Argument), since α extends to an α-isomorphism ∗ m m λ : L → L with λ(x) = x, the class Cα(p) = λ (Cp) = Cp . Hence T is a κ-rational ramiﬁcation type.

2. T is rigid so G = Gal(L/k(x)) has trivial center. Firstly, fix some α ∈ Gal(κ/κ). For each α ∈ Aut(k) there is an α-isomorphism λ : L → L0 where L0 is some finite Galois 0 0 1 extension of k(x). Let G = Gal(L /k(x)). For each p ∈ P = κ ∪ {∞} let Cp be the 0 0 class of G associated with p, and let Cp be the class of G associated with p. Now T is m κ-rational by hypothesis, therefore α(p) ∈ P , say α(p) = q, and so Cα(p) = Cq = Cp . So 0 ∗ m ∗ ∗ 0 by Theorem 4.3.11, Cq = λ (Cp ) = λ (Cq). Now λ is an isomorphism from G → G , 0 0 0 and it takes the class Cq to Cq. Therefore (G, P, Cp) ∼ (G ,P,Cp), and so L/k(x) and L0/k(x) have the same ramification type T . Now k = κ is an algebraically closed subfield of C by definition, T is a rigid type, so by Lemma 5.1.6 there is at most one finite Galois extension of this type up to k(x)-isomorphism. So L and L0 are k(x)-isomorphic. Let this 0 isomorphism be µ : L → L ,and define a = µλ : L → L. Then a|k(x) = µλ|k(x) = Id so a is an α-isomorphism. Recall from Lemma 5.1.9 that if for each α ∈ Gal(κ/κ) there exists and α-isomorphism λ : L → L with λ∗ = Id then L is defined over κ. Thus an α-isomorphism satisfying this property will complete the proof.

0 ∗ ∗ ∗ ∗ ∗ 0 0 Cq = λ (Cq) ⇒ a (Cq) = µ λ (Cq) = µ (Cq) = Cq as µ : L → L is a k(x)-isomorphism. ∗ Therefore a fixes all the classes Cq. By Lemma 4.4.6 then, a is an inner automorphism. Given h ∈ G, say a∗(h) = ghg−1 with g ∈ G. g ∈ G induces g∗ : G → G by h 7→ ghg−1. −1 ∗ ∗ −1 ∗ −1 Now define b = g a. Then b = (g ) a = Id. Moreover b|k(x) = g |k(x)a|k(x). But a is an α-isomorphism so is identity on k(x), and g ∈ G = Gal(L/k(x)) so fixes k(x) by definition of Galois group. Therefore b is an α-isomorphism with b∗ = Id, and this completes the proof.

38 A usefull observation is that T = [G, P, (Cp)p∈P ] is κ-rational if P ⊂ κ ∪ {∞} and each class Cp is rational in the usual sense. Indeed, if α ∈ Gal(κ/κ) and p ∈ P ⊂ K ∪ {∞} then α(p) = p m and so by rationality of classes Cp = Cp = Cα(p). Theorem 5.2.3. Given a rigid and κ-rational ramification type T = [G, P, C] there exists a unique finite Galois extension L/C(x) of this type, defined over a purely transcendental extension κ(t1, . . . , ts) of κ and G occurs regularly over κ.

Proof. T is rigid, so by 4.4.8, there exists a unique finite Galois extension of C(x) of this type, say L/C(x). L is defined over a finitely generated extension of κ, say κ1 = κ(a1, . . . , ar). If t1, . . . ts is a collection of elements among a1, . . . , ar maximal with respect to being algebraically independent, then κ1 is finitely generated and algebraic over κ0 = κ(t1, . . . , ts), so κ1 is finite over the purely transcendental extension κ0 of κ. Let k = κ0 = κ1. Then by Lemma 5.1.2 L is defined over k, and Lκ/κ(x) is of the same type as L/k(x) (Theorem 5.1.6), and this type is rigid and κ-rational by hypothesis.

Given an element g ∈ Gal(κ0/κ0), since κ ⊂ κ0, can restrict g to g ∈ Gal(κ/κ). Therefore T is κ-rational ⇒ T is κ0-rational. Hence Lκ has rigid and κ0-rational type. Hence by Theo- rem 5.2.2 Lκ is defined over κ0. Let M be the subfield of Lκ, regular of κ0, Galois over κ0(x), from the definition 5.1.1. Then M is also a subfield of L with the same properties ⇒ L is defined over κ0. Now by Lemma 5.1.3 ∼ G = Gal(L/C(x)) = Gal(M/κ0(x)) = Gal(M/κ(t1, . . . , ts, x))

M is regular over κ0 by deﬁnition, and κ0 is regular over κ, therefore M is regular over κ. That is, H occurs regularly over κ.

Recall that the Inverse Galois Problem seeks to realise groups as Galois groups of extensions of Q. With that in mind, consider the special case of the theorem, with κ = Q. This gives a criterion for realising groups over Q. Theorem 5.2.4. (Rational Rigidity Criterion) Let G be a group with a rigid tuple (C1,...,Cr) such that Ci are rational conjugacy classes. Then G occurs regularly over Q.

Proof. This is an immediate corollary of Theorem 5.2.3 with P ⊂ Q ∪ {∞}. In more complicated examples it is often difficult to find rigid tuples that are rational. Often one uses a field that provides a less strict κ-rationality condition. The best and most commonly ab ab used example of this is the field Q , generated by all roots of unity. Let α ∈ Gal(Q/Q ). ab Then α(ζn) = ζn. Thus if P ⊂ Q ∪ {∞} the α(p) = p and so Cα(p) = Cp trivially satisfies the ab definition of Q -rationality. Hence, ab Corollary 5.2.5. (Q Rigidity Criterion) ab Let G be a group with rigid tuple (C1,...,Cr). Then G occurs regularly over Q Recall that the definition of rational classes was extended to ramification types T in 5.2.1. Now generalise to define a κ-rational class vector C.

Deﬁnition 5.2.6. Let C = (C1,...,Cr) be a class vector in G, a ﬁnite group. Then C is m m κ-rational if C1 ,...,Cr is a permutation of C1,...,Cr for each m ∈ Z such that there exists −1 m α ∈ Gal(κ/κ) with α (ζn) = ζn where n = |G|

39 It would be preferable to have a general condition that applied to class vectors, rather than speciﬁcally on ramiﬁcation types as in Theorem 5.2.3. Hence

Theorem 5.2.7. (The General Rigidity Criterion) Let κ ⊂ C, and let G be a ﬁnite group. If C is a rigid and κ-rational class vector then G occurs regularly over κ.

Proof. Need to show that a κ-rational class vector ensures a κ-rational ramification type. Let −1 m Z = Gal(κ(ζn)/κ). If α ∈ Z then α (ζn) = ζn . Hence define an action of Z on the class m Z vector C by α(C) = C . Let Z1 be the stabiliser of C1 in C. Consider the quotient Z1 = {α1Z1, . . . , αsZ1}. Now α1(C1), . . . , αs(C1) are distinct. Moreover, by κ-rationality they are a subset of C. WLOG say αi(C1) = Ci (possibly relabelling some of the classes). Now Z1 ≤ Z = Gal(κ(ζn)/κ), so Z1 acts on κ(ζn). Consider the fixed field κ1 of this action. By properties of Galois groups, since Z1 ≤ Z, and Z fixes κ, then κ1/κ is an extension. Let p1 be a primitive element for the extension. Define pi = αi(p1). Then Z acts on p1, . . . , ps. Now the stabiliser of p1 is Z1, which is the stabiliser of C1 under Z. Hence there is an isomorphism pi 7→ Ci. m Further, Cα(p) = α(Cp) = Cp , as is required for a κ-rational ramification type. However, the set {p1, . . . , ps} needs to be extended to have size r = |C|. Do so by induction. Let P = {p1, . . . , pr}, and define T = [G, P, (Cp)p∈P ]. Then this is a κ-rational ramification type. Now by Theorem 5.2.3, G occurs regularly over κ.

Recall Hilbert’s Irreducibility Theorem, which said that Q is a hilbertian ﬁeld. Therefore, if G occurs regularly over Q, then G occurs as a Galois group over Q. This gives the most appropriate form of a rigidity criterion in the context of the Inverse Galois Problem over Q, this is the Rigidity criterion over Q.

Theorem 5.2.8. (Rigidity Criterion over Q) Let G be a ﬁnite group with rigid and rational class vector C. Then G occurs regularly over Q. Therefore G is a Galois group over Q, and the Inverse Galois Problem has positive solution for G.

40 Chapter 6

Applications of Rigidity

In this chapter, the results obtained to this point are applied to show that the Inverse Galois Problem has a positive solution in these examples. Rigidity is applied to the symmetric and alternating groups on n letter. Although Sn has already been shown to have a positive solution to the Inverse Galois Problem as an application of Hilbert’s Irreducibility Theorem, it is presented again as a nice example of how rigidity can be used. Then, by imposing a congruence condition on p, PSL2(p) is realised as a Galois group over Q. The value of the character theoretic formula is then highlighted with several examples. It is particularly useful when approaching the sporadic groups, as their character tables are known. To demonstrate this M12 and the Monster are realised. As was alluded to earlier, it is often difficult to ensusre rationality of the class vector, and as a result, often one has to settle for a realisation over a different field. An ab example of such a case is presented as the Ree groups in characteristic 3 are realised over Q .

6.1 Sn and An as Galois groups over Q

In Sn the cycle shape determines the conjugacy class. Now conjugation by a given cycle on an element of the same cycle shape does not change the cycle shape, therefore the conjugacy classes of Sn are rational. It remains to ﬁnd a rigid class vector. Then by Theorem 5.2.8 Sn will occur as a Galois group over Q. I will construct such a rigid tuple.

Let a = (1 ··· n − 1) and b = (n − 1 n n − 2 ··· 1) be an (n − 1)-cycle and an n-cycle. Now let (i) (n−1) (n) C denote the conjugacy class of i-cycles in Sn. Then a ∈ C , b ∈ C .

Claim 1: A subgroup of Sn containing a and b is transitive. Pf: Is clear. a takes x 7→ x + 1 for x ∈ [1, . . . , n − 2], b takes x 7→ x − 1 for x ∈ [2, . . . , n − 2] and takes n − 1 7→ n. So by repeated application of a and/or b one can take any x ∈ [1, . . . , n] to any other. Thus H is indeed transitive.

Claim 2: A transitive subgroup of Sn, with n ≥ 2 which contains an (n − 1)-cycle and a transposition, is the whole of Sn. Pf: Let H be said subgroup. Without loss of generality we may assume the (n − 1)-cycle is a = (1 ··· n − 1) (possibly after relabelling the elements). Let (i j) be the transposition. Since H is transitive on [1, . . . , n] by hypothesis, there exists h ∈ H such that h(j) = n. Moreover, the element (i j)h = h(i j)h−1 = (hi hj) = (x n) ∈ H by closure. Now am(x n)a−m = (am(x) am(n)). Nut a ﬁxes n, so (am(x) am(n)) = (x + m n) where m ∈ [1, . . . , n] and where x + m cycles through (1 ··· n − 1). Therefore by repeated application of a to (x n) we obtain (1 n), (2 n),..., (n − 1, n) ∈ H. Finally, (i n)(j n)(i n) = (i j) ∈ H. Therefore H contains all

41 transposition and thus H = Sn.

(2) (n−1) (n) (2) (n−1) Let C = (C ,C ,C ) and let σ1 = (n − 1 n) ∈ C , σ2 = (1 ··· n − 1) ∈ C , σ3 = (n) (n − 1 n n − 2 ··· 1) ∈ C . Then the subgroup generated by σ = (σ1, σ2, σ3) is the whole of Sn by the two claims. Moreover,

σ1σ2σ3 = (n − 1 n)(1 ··· n − 1)(n n − 1 n − 2 ··· 1) = (n − 1 n)(n − 1 n) = 1 0 Therefore σ ∈ C generates Sn and satisﬁes σ1σ2σ3 = 1. It remains to show that if σ ∈ C is 0 another triple with the same properties then there exists τ ∈ Sn such that τσiτ = σi.

Claim 3: Let (C1,C2,C3) be a class vector in H with generators g1, g2, g3 such that gi ∈ Ci and 0 0 −1 0 g1g2g3 = 1. If H has trivial center and for each g2 ∈ C2 with (g1g2) ∈ C3 and H = hg1, g2i −1 0 −1 there is h ∈ H with hg1h = g1 and hg2h = g2, then (C1,C2,C3) is a rigid class vector. Pf: Clear from deﬁnition of rigidity.

0 (1) 0 (n) 0 Now let σ1 ∈ C . If σ2σ1 ∈ C then σ1, σ2 generate Sn by Claim 1 and Claim 2. This 0 0 0 0 0 σ1 is the g2 in Claim 3. Well if σ2σ2 is an n-cycle then σ2 must be of the form σ2 = (i n). Now −1 σ2(i n)σ2 = (σ2(i) σ2(n)) = (i + 1 n) as in the proof of claim 2. Therefore repeated conjugation by σ2 will map (i n) to (n−1 n) = σ1. 0 That is to say that σ1 conjugates to σ1. Now it is clear tht σ2 commutes with σ2 and to whatever 0 the power of σ2 that conjugates σ1 to σ1 will conjugate σ2 to itself. Thus the condition in Claim 3 is satisﬁed, and hence the class vector is rigid.

(2) (n−1) (n) Therefore the class vector C = (C ,C ,C ) is a rigid and rational class vector in Sn, and hence by Theorem 5.2.8, Sn occurs as a Galois group over Q.

I now appeal to a result of Serre to realise An over Q. For the proof I follow [12].

Lemma 6.1.1. Let G be a group with rigid and rational class vector C = (C1,C2,C3). Then if H ≤ G such that |G : H| = 2 then H occurs regularly over Q.

Proof. First construct an appropriate ramification type T . Let P = {p1, p2, p3} where pi ∈ Q are distinct. Then T = [G, P, C] is a rigid and rational ramification type. Now Theorem 5.2.3 says there exists a finite extension M/C(x) of type T , moreover G occurs regularly over Q. That is, M/C(x) is defined over a purely transcendental extension κ0 = Q(t1, . . . , ts), so by definition there exists subfield M0 := Mκ0 regular over κ0, Galois over C(x) and such that |M0 : κ0(x)| = |M/C(x)|. And by Lemma 5.1.3 Gal(M0/κ0(x)) = Gal(M/C(x)) = G. Fix such an isomorphism. Now a subgroup H ≤ G of index 2 corresponds to√ a quadratic extension N0 of κ0(x) inside M0 by the Galois correspondence. Say N0 = κ0(x)( f) with f ∈ κ0[x] a non- constant polynomial√ with no multiple irreducible factors and hence no multiple roots over C. Let N = C(x)( f). Then N ⊂ M is a quadratic extension of C(x) by Lemma 3.1.4. By Lemma 4.3.7 the ramification index e of points from N/C(x) and M/C(x) are the same. Therefore the branch points of N/C(x) are branch points of M/C(x). Now the branch points of M/C(x) are Q p1, p2, p3, by construction of the type T . Now we may take f of the form (x − pi), where pi is a branch point of N/C(x), and ∞ is a branch point if and only if the number of branch points is odd [12]. That is to say that there is always an even number of branch points for quadratic extensions. Assume without loss of generality that p1, p2 are the branch points of N/C(x). Then f(x) = b(x − p1)(x − p2) with b ∈ κ0 Thus √ . p f N0 = κ0( f) = κ0 (x − p2) (6.1)

42 2 c(x−p1) 2 Let z = . Then (6.1)= κ0(x)(z). Finally, notice that x ∈ κ0(z) since z is a linear (x−p2) function of x. Therefore N0 = κ0(x)(z) = κ0(z) is a rational function ﬁeld. Therefore H = Gal(M0/N0) = Gal(M0/κ0(z)) and so H occurs regularly over κ0 and therefore also occurs regulary over Q by Theorem 5.2.3.

We are now able to realise An as a Galois group over Q as a corollary to this, and given that |Sn : An| = 2.

6.2 PSL2(q) for q 6≡ ±1 mod 24 as a Galois group over Q A New Rigidity Criterion For this example it is convenient to give a new rigidity criterion that applies to groups modulo their center. Let C = (C1,...,Cr) be a class vector in a group H. Recall, we say C is weakly 0 0 rigid in H if ∃ generators σ1, . . . , σr of H with σ1 ··· σr = 1 and σi ∈ Ci such that if σ1, . . . , σr 0 is another generating set with the same property then ∃ γ ∈ Aut(H) such that γ(σi) = σi. Deﬁnition 6.2.1. We say C is quasi-rigid in H if it is weakly rigid and the automorphism in the deﬁnition is inner i.e. γ ∈ Inn(H).

For a rigid class vector we require the existence of a unique element h ∈ H such that −1 0 hσih = σi. This uniqueness is equivalent to saying that H has trivial center. For a weakly- rigid class vector, since γ is deﬁned on a generating set, then it must be unique. Thus in the rigid case any automorphism of H ﬁxing each Ci is inner. Therefore C is quasi-rigid and has trivial center ⇐⇒ C is rigid.

Lemma 6.2.2. Let k = κ and let L/k(x) be an FG-extension with Galois group G. Let L0 be the fixed field of Z(G). If the ramification type of L is quasi-rigid and κ-rational then L0 is defined over κ.

Proof. As in the proof of Theorem 5.2.2(2), for each α ∈ Gal(κ/κ) we get an α-automorphism ∗ λ of L with λ = Id. Let θ generate L over k(x). Let θ1, . . . , θt be the conjugates of θ over κ(x). Then θi ∈ L since L is Galois over κ(x). Let L1 := κ(x)(θ1, . . . , θt). Then L1 is an FG-extension of κ(x). Let κ1 be the algebraic closure of κ in L1. Then L is deﬁned over κ1 ∼ as in Lemma 5.1.4, and G1 = Gal(L1/κ1(x)) = G. Again, as in proof of Lemma 5.1.4, G1 is C normal in H := Gal(L1/κ(x)). Let C = CH (G1), then H = G1C. Now L1 is Galois over κ(x) with Galois group . . H (G1C) ∼ G1 G1 /C = /C = (G1 ∩ C) = Z(G1)

C C where here L1 denotes the fixed field of C in L1. We have that κ(x) = κ1(x) ∩ L1 follows C C from H = G1C. Thus L1 is regular over κ. Furthermore, from Lemma 5.1.3, we have L1 ⊂ Z(G1) Z(G) 0 0 0 C L1 ⊂ L = L . Thus L is defined over κ with Lκ = L1 . If we now replace Theorem 5.2.2 with this lemma in the proof of the Theorem 5.2.7 we get . Theorem 6.2.3. If C is a quasi-rigid and κ-rational class vector in H then H Z(H) occurs regularly over κ.

43 The Groups PSL2(q) as Galois Groups ab We consider only p odd and ﬁrstly show that PSL2(q) occurs regularly over Q for powers of odd primes. Then, by imposing a condition on the prime p we show PSL2(p) occurs regularly over Q.

Let q be an odd prime. 1 1 Lemma 6.2.4. Let U = . Let U ∈ SL (q) of trace 2, not upper triangular. Then 1 0 1 2 1 0 ∃ A ∈ SL (q) such that A ∈ C (U ) and A conjugates U into a matrix of the form 2 SL2(q) 1 c 1

a b Proof. Let q odd prime. Consider U = ∈ SL (q) not upper triancular, so c 6= 0. c d 2 1 m Take A of the form A = , which clearly centralises U . Then 0 1 1

a + mc ∗ AUA−1 = c ∗

1 ∗ With m = c−1(1 − a) get AUA−1 = . But, trace(U) = 2 ⇒ trace(AUA−1) = 2 and c ∗ −1 −1 AUA ∈ SL2(q) ⇒ det(AUA ) = 1, thus

1 0 AUA−1 = c 1

Lemma 6.2.5. There are precisely two conjugacy classes of non-identity elements of SL2(q) with trace 2, say C1,C2, where C1 is the class containing U1. 1 u 1 0 Proof. Non-identity elements in SL (q) of trace 2 are of the form , where 2 0 1 v 1 u, v 6= 0 in Fq. I claim these matrices are in C1 if and only if u or −v are square in Fq, respectively. Indeed, let U ∈ SL2(q) of trace 2. By Lemma 6.2.4, if U is not upper triangular 1 0 then it is conjugate to some . We have c 1

0 1 1 0 0 1 −1 1 −v = (6.2) −1 0 v 1 −1 0 0 1

1 u So we have that U is conjugate to some (note: if U is upper triangular then this is 0 1 trivial). Need to see when a matrix of this form is conjugate to U1. Suppose B ∈ SL2(q) is such that 1 u BU B−1 = 1 0 1

U1 has eigenvalue 1 with corresponding eigenspace E1 = Sp{(1, 0)}. Then B must be upper triangular as it ﬁxes this eigenspace. Combining this with the fact that det(B) = 1 we see that

44 w ∗ B is of the form . Now 0 w−1

w ∗ w ∗ −1 1 w2 U = 0 w−1 1 0 w−1 0 1

Hence by 6.2 we have proved the claim, and the lemma follows.

We state a classical result of Dickson

(Dickson) Let q be a power of the odd prime p and Fq = Fp(c) with c 6= 0. Then the 1 1 1 0 matrices and generate SL (q) unless q = 9. 0 1 c 1 2

Lemma 6.2.6. Let q be a power of an odd prime p with q 6= 9. Let Fq = Fp(τ) with τ 6= 2. Let τ − 1 1 −1 C(τ) be the class of SL (q) containing σ := . 2 3 τ − 2 1

1. If 2−τ is nonsquare in Fq then the class vector C = (C1,C2,C(τ)) is quasi-rigid in SL2(q).

2. If 2 − τ is a square in Fq then the class vector C = (C1,C1,C(τ)) is quasi-rigid in SL2(q). 1 0 Proof. Let σ := U . Consider σ := . If 2 − τ is nonsquare in then this is in 1 1 2 τ − 2 1 Fq C2 and if 2 − τ is square in Fq then it is in C1 by Lemma 6.2.5. Now

1 1 1 0 τ − 1 1 = 0 1 τ − 2 1 τ − 2 1 shows that ∃ a triple σ = (σ1, σ2, σ3) ∈ C such that σ1σ2σ3 = 1. By Dickson’s theorem σ1 0 0 0 and σ2 generate SL2(q). It remains to show that for any other triple (σ1, σ2, σ3) with the same 0 0 properties ∃ γ ∈ Aut(SL2(q)) such that γ(σi) = σi and that γ is inner. If σ1 = σ1 then the inner automorphism is conjugation by an element of Γ := CSL2(q)(σ1). So to show quasi-rigid 0 0 need to show that all σ2 ∈ C2 when 2 − τ is nonsquare, or all σ2 ∈ C1 when 2 − τ is a square, 0 0 with σ1σ2 of trace τ such that hσ1, σ2i = SL2(q) are conjugate under Γ. 0 0 Now σ1 is upper triangular, and hσ1, σ2i = SL2(q), so σ2 is not upper triangular. Hence, by 1 0 Lemma 6.2.4, σ0 is conjugate to under Γ. Thus 2 c 1

1 0 trace(σ σ0 ) = τ ⇒ trace σ = τ 1 2 1 c 1

But 1 0 1 1 1 0 1 + c 1 σ = = 1 c 1 0 1 c 1 c 1 1 0 has trace τ ⇒ c = τ − 2. That is σ0 is conjugate to under Γ, as required. 2 τ − 2 1

ab Corollary 6.2.7. PSL2(q) occurs regularly over Q for each odd prime power q.

45 . Proof. PSL2(q) = SL2(q) {±1}. We have shown the existance of a quasi-rigid class vector ab (C1,C2,C(τ)) or (C1,C1,C(τ)). Since each tuple of conjugacy classes in Q -rational we can apply Theorem 6.2.3, since Z(SL2(q)) = {±1}. For the exceptional case, q = 9 we know that ∼ PSL2(9) = A6, and we have demonstrated a stronger result earlier for this, namely that A6 occurs regularly over Q.

Theorem 6.2.8. The group PSL2(p) occurs regularly over Q for each prime p 6≡ ±1 mod 24 ∼ ∼ Proof. Firstly, PSL2(2) = S3 and PSL2(3) = A4 occur regularly over Q as demonstrated in Section 6.1. Now consider p > 3. Let τ ∈ Fp such that τ 6= 2 and τ is not a square in Fp. Then by Lemma 6.2.6 the class vector C = (C1,C2,C(τ)) is quasi-rigid in SL2(p). Let m be an m m integer prime to p. Then (C1 ,C2 ) = (C1,C2) or (C2,C1) depending on whether m is square mod p or not. Thus if C(τ) is rational then C is a quasi-rigid, rational class vector, which implies, by Theorem 6.2.3, that PSL2(p) occurs regularly over Q.

It remains to ﬁnd τ ∈ Fp with 2 − τ nonsquare in Fp and C(τ) rational. Here C(τ) is the class consisting of all elements of trace τ. If 2 is nonsquare mod p then set τ = 0. Then 2 − τ = 2 is non-square by assumption. Now C(0) consists of elements of order 4, and these are conjugate to their inverse. Thus C(0) is rational. If 3 is nonsquare mod p then set τ = −1. Then 2 − τ = 3 is non-square by assumption. Now C(−1) consists of elements of order 3 and these too are conjugate to their inverse and thus C(−1) is rational.

But 2, 3 nonsquare mod p is equivalent to p 6≡ ±1 mod 24 by elementary number theory.

6.3 M12 as Galois group over Q Another example of the power of the rigidity method is that it has been used to realise 24 of the 26 so called sporadic groups over Q. Using Corollary 4.5.7 the quest to find rigid and rational class vectors in a group G, and hence to realise G as a Galois group over Q, can be reduced to examination of the character table of G, most useful given Conway’s Atlas of finite groups [1]. An example is presented in which the character table of the Mathieu group M12, a sporadic group, can be used to show that M12 occurs over Q. In most applications of the rigidity criteria to sporadic groups, and in general, first the character tables are used to verify the formula 4.5.7 holds for a given class vector C, and then it remains to show that a given system in C generates the group. This is usually done by showing that H = hσi cannot be contained in any maximal subgroup of the group [7]

Recall Σ(b C) = {σ ∈ C : σi ∈ Ci, σ1 ··· σr = 1}, where C = (C1,...,Cr). C is rigid if there exists σ ∈ Σ(b C) such that G = hσi and if σ0 ∈ Σ(b C) is another tuple with the same −1 0 properties then there is g ∈ G such that gσig = σi for i = 1, . . . , r. That is to say that G acts transitively on Σ(b C) by conjugation if C forms a rigid class vector. Then Corollary 4.5.7 says this second condition is equivalent to σ satisfying the formula

X χ(σ1) ··· χ(σr) |CG(σ1| · · · |CG(σr)|) = (6.3) χ(1)r−2 |G|r−2|Z(G)| χ∈Irr(G)

46 From [4] the size of Σ(b C) is calculated to be r−1 |G| X χ(σ1) ··· χ(σr) |Σ(b C)| = r−2 (6.4) |CG(σ1)| · · · |CG(σr)| χ(1) Irr(G)

From which, if Z(G) = 1 and |Σ(b C)| = |G| then rearranging gives (6.3). That is to say, if |Σ(b C)| = |G| and Z(G) = 1 then C is a rigid tuple if there is σ ∈ C generating G. This can be used to ﬁnd so called candidates for rigid class vectors, given that the classes are rational.

In the vast majority of applications of the rigidity method, rigid tuples consist of three conjugacy classes. I therefore restrict my attention to rigid triples in the arguments that follow.

Before consulting the Atlas [1] to see if one can find a σ ∈ Σ(b C) generating G and satisfying (6.3), the following lemma helps by eliminating some possible candidate tuples. Lemma 6.3.1. If Out(G) 6= 1 and α ∈ Out(G) fixes C then C cannot be a rigid tuple. Proof. Suppose C is a rigid class vector fixed by an α ∈ Out(G). Then C rigid ⇒ there exists 0 σ = (σ1, σ2, σ3) ∈ Σ(b C) such that G = hσi and if σ ∈ Σ(b C) has the same properties then there −1 0 exists g ∈ G such that gσig = σi, i.e. G acts transitively on Σ(b C) by conjugation.

α α α α Now σ1σ2σ3 = 1 ⇒ σ1 σ2 σ3 = 1, and σi ∈ Ci since α ﬁxes C by hypothesis. Therefore α α α α σ = (σ1 , σ2 , σ3 ) ∈ Σ(b C). But G is transitive on Σ(b C), therefore hσi is centralised by an outer automorphism, and so doesn’t generate G. This is a contradiction.

Hence if Out(G) 6= 1 then we need only consider class vectors not fixed by Out(G). Now Out(M12) = 2 [1]. The classes not fixed by Out(M12) are 4A and 8A, see Appendix B for character table [4]. There are 7 possible candidates for rigid class vectors containing one of these classes they are (2A, 3A, 8A), (2A, 3B, 8A), (2A, 4A, 6A), (2B, 3A, 8A), (2B, 4A, 10A), (3A, 3B, 4A) and (3B, 4A, 4A) 1. It remains to find a candidate class vector that can generate G. However, as can be seen from the character tables of the maximal subgroups of M12 in Appendix B, all of these triples generate a proper subgroup of G, and hence are not rigid 2.

Now consider the extension by the full outer automorphism group, M12.2. This contains a candidate triple (2A, 4C, 12A) which doesn’t generate a proper subgroup since no maximal subgroup of M12.2 contains representatives of both classes 4C and 12A. To see this notice that ψi(4C) = 0 for i = 1,..., 8, 11, 14 and ψi(12A) = 0 for i = 1, 2, 3, 4, 6, 7, 9, 10, 12, 13, so at least one is alway 0 (Appendix B, Figure B.1).

Finally, use Hilbert’s result Lemma 6.1.1 to see that, since |M12.2 : M12| = 2, and we just demonstrated that M12.2 contains a rigid triple of rational classes, then M12 occurs regularly over Q.

2 2 ab 6.4 The Ree Groups G2(q ) as Galois groups over Q The exceptional groups of Lie type consist of ten families of ﬁnite simple groups. Unlike the classical Lie groups, they do not posses a convenient matrix representation. Currently there is

1All the information required to calculate |Σ(b C)| is available in Appendix B. A computer can easily check all possible triples including at least one of 4A, 8A to establish these candidates. However, I got these from [4] 2 To see this notice that for every candidate triple (C1,C2,C3) the permutation character of the classes of maximal subgroups ψi(C1), ψi(C2), ψi(C3) 6= 0 for some i ∈ [1,..., 14]

47 no uniform way of realising all the exceptional groups of Lie type, infact it is not yet known whether they can all be realised as Galois groups over Q [7]. Rigidity has oﬀered the most fruitful results in this area, the problem being that it is ofen diﬃcult to satisfy the rationality condition on rigid class vectors. As discussed earlier, when this is the case, one often uses the ab Q rigidity criterion. The following is an example gives a realisation of the Ree groups in ab characteristic 3 over Q .

A useful result when applying the rigidity criterion follows. It is helpful when showing that a system is not contained in a maximal subgroup of the group in question, and hence is used to show that a given system can generate the group. It appears in many applications of rigidity (for instance [4], Lemma 1).

a b c Lemma 6.4.1. Suppose G = hσ1, σ2, σ3 : σ1 = σ2 = σ3 = σ1σ2σ3 = 1i where a, b, c are pairwise coprime. Then G = [G, G] = hxiG . Proof. Consider the quotient G [G, G]. Then o(σ1) divides a = o(σ1) and bc = o(σ2)o(σ3). . Therefore σ1 is trivial, as a, b, c pairwise coprime. Similarly, σ2 and σ3 are trivial in G [G, G]. G 0 0 G 0 b 0 c Let H = hσ1i . Let σ2, σ3 be the images of σ2, σ3 in /H . Then (σ2) = 1, (σ3) = 1 and 0 0 σ2σ3 = 1 hence H = G. 2 2 2 2m+1 The conjugacy classes of G = G2(q ) in characteristic 3 (let q = 3 ) and the relevant portion of the character table were determined in [13]. A complete list of maximal subgroups 2 2m+1 was determined in [6]. The character table of G2(3 ) is presented, with the same notation used in [13], where ξi are irreducible characters, C2 is the unique class of involutions, C3 is the the class of 3-elements lying central in a√ Sylow 3-subgroup and Cs is one of the classes 2 containing semi-simple elements of order q − 3q + 1 ∈ Z.

1 C2 C3 Cs ξ 1 1 1 1 1 √ √ √ ξ 3 q(q2 − 1)(q2 + 3q + 1) − 1 (q2 − 1) − 1 (3q2 + 3q) −1 5 √6 √ 2 6 √ 3 2 2 1 2 1 2 ξ7 6 q(q − 1)(q + 3q + 1) − 2 (q − 1) − 6 (3q + 3q) −1

Lemma 6.4.2. The class vector C = (C2,C3,Cs) is rigid

Proof. This portion of the character table is the relevant portion as ξ1, ξ5, ξ7 are the only characters that do not vanish anywhere on C2,C3,Cs, and hence are the only classes that contribute to the formula in Corollary 4.5.7. Let σ = (σ1, σ2, σ3) with σ1 ∈ C2, σ2 ∈ C3, σ3 ∈ Cs. 2 2 6 2 6 Given that | G2(q )| = q (q − 1)(q + 1), can calculate,

2 4 |CG(σ1)| = q (q − 1) 6 |CG(σ2)| = q √ 2 |CG(σ3)| = q − 3q + 1 with the class sizes from [13]. Now plugging the numbers in one veriﬁes that the following holds true, X χ(σ1) ··· χ(σs) |CG(σ1)| · · · |CG(σs)| = χ(1)s−2 |G|s−2|Z(G)| χ∈Irr(G)

48 It remains to show that hσi generates G. Let H := hσi. The maximal subgroups of G according to Kliedman in [6] are

2 L2(q ) × 2 2 (2 × D q2+1 ).3 2 [q6].(q2 − 1) √ (q2 + 3q + 1).6 √ (q2 − 3q + 1).6 2m + 1 2G (r2) with r2 = 32k+1 where is prime 2 2k + 1

Now the orders of σi are pairwise coprime, so by Lemma 6.4.1, H = [H,H]. Now the only 2 2 2 2 non-solvable maximal subgroups are L2(q ) × 2 and G2(r ) [7]. But the order of L2(q ) × 2 is prime to the order of σ3, which would contradict Lagrange. So the only possible maximal 2 2 subgroup that H√ could be contained in is G2(r ). But the largest order of√ an element in 2 2 2 2 2 1 G2(r ) is r + 3r + 1 ≤ 2r + 1 which is less than the order of σ3 = q − 3q + 1 ≥ 2q2+1 [7]. Therefore H is not contained in a any maximal subgroup, and so H = G. That is, σ is a generating system for G. Therefore, by Corollary 4.5.7, C is a rigid class vector.

Finally it remains to check when C forms a rational class vector, whence application of the Rigidity Criterion for Q would give that G occurs as a Galois group over Q. Now C2 and C3 are rational classes. But Cs is not. Thus one cannot use the rigidity criterion over Q and so the ab best result from this class vector is obtained using the rigidity criterion over Q , and provides ab the result that G occurs as a Galois group over Q .

6.5 Realising the Monster over Q The Monster group M, or the Fischer-Griess Monster is the largest sporadic group, with order

246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71 and represents one of the most celebrated results in the classification of finite simple groups [1]. There is a famous paper by Thompson, [11] in which he introduces rigidity and uses to show that the Monster group occurs as a Galois group over Q. This chapter concludes with Thompson’s realisation of M, a magnificent example of the power of rigidity; imagine trying to find a polynomial whose Galois group is M by hand! But one does exist.

M has no outer automorphisms, and contains two classes of elements of order two and three classes of elements of order 3 [1]. Using a computer and the ATLAS [1] one can check possible candidates for rigid tuples, as in the example of M12. These candidates are triples that satisfy the formula in Corollary 4.5.7(2). If there are any, it remains to show that M is generated by one of them. The aforementioned computation shows that there are candidates. Consider, as Thompson does in [11], such a candidate C = (2A, 3B, 29A).

Let H = hσ1, σ2, σ3 : σ1 ∈ 2A, σ2 ∈ 3B, σ3 ∈ 29A, σ1σ2σ3 = 1i. Now 2, 3, 29 are pairwise coprime, hence by Lemma 6.4.1, H = [H,H]. Let H = H /L where L is a maximal normal

49 subgroup of H. By the classiﬁcation of ﬁnite simple groups the possibilities for H are

L2(29)

L2(59) Rv 0 F i24 since 29||H| [4]. The involutions in the class 2A are {3, 4, 5, 6}-transposition. That is for x, y ∈ 2A, o(xy) ≤ 6, but the classes of involutions in L2(29),L2(59) and Rv are not {3, 4, 5, 6}- 0 transpositions [4]. Now suppose H = F i24. This is a contradiction as the character 196883a of 0 0 M (see [1], pg.220) does not restrict to a character in F i24 [4]. Therefore H is not F i24. So 0 if H = F i24 then L 6= 1. Now for x ∈ 29A, |CM(x)| = 3 · 29 from [1]. From [4], if E is an elementary abelian 2-group (cyclic and every element order 2) then |E| ≤ 224, o(2) = 28, o(3) = 28, o(5) = 14, o(7) = 7( mod 29). Putting this together with the classification of finite simple 0 0 groups shows that H = 3 · F24 or M [4]. Suppose that H = 3 · F24. Then H = CM(u) for u ∈ 3A. Let σ1 ∈ 2A, σ2 ∈ 3B, σ3 ∈ 29A such that σ1σ2σ3 = 1. From this create another −1 generating set of a different class vector, σ1σ2(uu )σ3 = 1 so consider

0 0 0 −1 σ1 = σ1 σ2 = σ2u σ3 = u σ3

−1 Since u ∈ Z(H) o(u σ3) = 87. Also, o(σ2u) = 3. But, consideration of candidate triples by evaluation of [1] shows that

|Σ(2b A, 3A, 87A)| = 0 |Σ(2b A, 3B, 87A)| = 0 as in the example of M12, and as defined in Section 4.5. The only possible triple is thus 0 (2A, 3C, 87A), i.e. σ2 = σ2u ∈ 3C. Now by the classification of finite simple groups hσ2, ui 0 contains at least 2 Suzuki element, 2 Fischer elements and 2 Thompson elements [4]. CM(σ2) = 3 × E. Now the character of 196883a of M restricts to 3 × E as

(1 + ω + ω) ⊗ 1a + (ω + ω) ⊗ 248a + (1 + ω + ω) ⊗ 4123a + 1 ⊗ 306281 + (ωω) ⊗ 61256a from [4]. For x ∈ 3A, 3B, 3C, 196883a(x) = 728, 53, −1 respectively. Thus the following classes are identified under this restriction, 3A in E with 3A in M, 3B and 3C in E with 3B in M. But 196883a(ux) = −1 for x ∈ 3A, 3B or 3C in E ⇒ ux ∈ 3C in M [4]. Therefore every elementary alelian subgroup of order 9 in M contains 0 or 6 Thompson elements [4], and thus 0 H is not 3 · F24 Therefore H = M and so condition 1 of Corollary 4.5.7 has also been satisfied. Therefore (2A, 3B, 29A) is a rationally rigid class vector in the Monster group, and thus there exists a finite Galois extension of Q with Galois group equal to M.

50 Chapter 7

Conclusion

The aim of this project was to explore the concept of rigidity, and to understand how it is used to address the Inverse Galois Problem. Suﬃcient background was developed to understand Riemann’s Existence Theorem and Hilbert’s Irreducibility Theorem. Rigidity was then deﬁned and the role it plays set out. The summary of ideas is, given a group G with a rigid tuple, Riemann’s Existence Theorem realises G over C(x). If the tuple is rationally rigid one can then descend to realise the group over Q(x), which, when combined with Hilbert’s Irreducibility Theorem guarantees the existence of an extension of Q with Galois group G. That is, G has a positive solution to the Inverse Galois Problem.

By considering a variety of examples, the application of these ideas in practice is demonstrated. In the introduction a motivating example was considered, as D8 was realised as a Galois group over Q by considering a well chosen polynomial. Looking back that example helps highlight the power of these ideas; attempting to ﬁnd a polynomial whose Galois group is the Monster, for instance, is completely unfeasible, but given that the character table is known, it was possible to show that it does have a positive solution to the Inverse Galois Problem.

To this day, rigidity has been used, much in the same way as demonstrated in the examples of this project, to realise a large number of groups over Q, including linear groups Ln(p), unitary groups Un(p), symplectic groups S2n(p) and orthogonal groups O2n+1(p) with certain congruence conditions on p, many exceptional groups of Lie type and all but two of the sporadic groups.

That said, the story doesn’t end here as the Inverse Galois Problem is still open. One area of focus is the extension problem. It seeks to use the composition series of a group to yield a tower of extensions whose relative Galois groups are simple. The problem arises when trying to ﬁt the extensions together in an appropriate way. The solvable groups are the easiest example of this, and Shafaravich was able to realise them all over Q using this idea [12]. However, this has not yet been successfully generalised to non-solvable groups.

This project has looked at some of the most fruitful techniques used in considering the In- verse Galois Problem, and has demonstrated how they have been applied. Probably the most famous application of rigidity is Thompson’s realisation of the Monster over the rationals, and it is in showing this that the aim of the project is realised.

51 Bibliography

[1] Conway, J. H., Curtis, R. T., Norton, S. P., Parker, R. A. and Wilson, R. A. [1985], An Atlas of Finite Groups, Oxford Univeristy Press, Oxford.

[2] Fraleigh, J. B. [1968], A First Course in Abstract Algebra, Addison-Wesley Publishing Company, London.

[3] Hatcher, A. [2002], Algebraic Topology, Cambridge University Press, Cambridge.

[4] Hunt, D. [1986], ‘Rational rigidity and the sporadic groups’, Journal of Algebra 99, 577– 592.

[5] Jacobson, N. [1985], Basic Algebra I, Freeman, New York.

[6] Kleidman, P. B. [1988], ‘The maximal subgroups of the chevally groups G2(q) with q odd, 2 the Ree groups G2(q) and of their automorphism groups’, Journal of Algebra 117, 30–71. [7] Malle, G. and Matzat, B. H. [1999], Inverse Galois Theorey, Springer, New York.

[8] Queen, C. S. [1996], ‘Factorial domains’, Proceedings of the American Mathematical Society 124(1).

[9] Rotman, J. J. [1965], An Introdution to the Theory of Groups, Allyn and Bacon, Cambridge.

[10] Rotman, J. J. [1998], Galois Theory, Springer, New York.

[11] Thompson, J. G. [1984], ‘Some ﬁnite groups that appear as Gal(L/K), where K ⊂ Q(µn)’, Journal of Algebra 89, 437–499.

[12] Volklein, H. [1996], Groups as Galois Groups, Cambridge University Press, Cambridge.

[13] Ward, H. N. [1966], ‘On Ree’s series of simple groups’, Trans. Amer. Math. Soc. 121, 62–89.

[14] Zannier, U. [2009], ‘On the Hilbert Irreducibility Theorem’, Rend. Sem. Mat. Univ. Pol. Torino 67(1).

52 Appendix A

GAP Code

GAP is a system designed for use in computational group theory. As well as a programming language it has a large database of algebraic objects. I have used GAP as one way of obtaining the character table of speciﬁc groups. I present the code, in red, together with the GAP output, in green.

GAP attempts to present the character table in a similar way to the ATLAS [1]. The first rows list, on the left, the primes dividing the order of the group, and then for each class the exponents of the prime factorisation of the centraliser order. The next row is the name of the conjugacy classes, and, where applicable, the further rows give all the primes p dividing the group order. Then the character table is presented, with irreducible character on the left ar- ranged according to degree. Zeros in the table are represented by a ‘·’. Algebraic irrationalities are denoted by a letter, and these are defined below (where applicable). The final computation is the size of the conjugacy classes.

53 Figure A.1: Character Table for SL2(8)

Figure A.2: Character Table for S4

54 Appendix B

Characters for M12

The following table from [4] displays the character table of M12, this is the 15 × 15 upper left table. The remaining 9 columns give the characters for M12.2. Following the convention from [1] the column headings are ﬁrstly the size of the conjugacy class, then the power maps, and then the name of the class. The irreducible characters are χi with i = 1,..., 15. The lower 14 × 21 table gives the permutation characters ψi with i = 1,..., 14, for the conjugacy classes of maximal subgroups.

55 Figure B.1: Character Table of M12 with additional information